Borel's Proof
The relevant passage from Borel occurs on pages 5152 of “Sur quelques points de la théorie des fonctions” [5]. Here, he stated that he had found an easy lemma, but that he would prove it nonetheless because it appeared to be interesting.
One can consider this lemma to be nearly evident; nevertheless, because of its importance, I am going to give a demonstration resting on a theory interesting by itself; there exist other simpler demonstrations. Here is the theorem: If one has on a straight line an infinite number of partial intervals, such that any point on the line is interior to at least one of the intervals, one can effectively determine a LIMITED NUMBER of intervals chosen among the given intervals and having the same property (any point on the line is interior to at least one of them).
Notice that Borel mentioned he would go out of his way to provide a proof using a technique that was interesting, rather than using the simplest method. As we work through the proof, we will see that Borel used the monotone convergence characterization of completeness. We should also point out that Borel made a few assumptions that are not explicit. These assumptions will appear in later proofs as well:

By “straight line,” Borel referred to an interval containing its endpoints  in other words, a closed and bounded interval. Students should be able to find counterexamples if this hypothesis is omitted.

By “partial intervals,” he meant “open intervals.”

By “LIMITED NUMBER,” he meant “finite number.”
Borel started by stating his assumption that the cover be countable.
It is as a matter of course that the word interior is always taken in the limited sense which excludes the extremities; it is easy to assure that, without this, the theorem would not be true. One can demonstrate directly that any point on the line is necessarily within the interior of a finite number (supposing the numbered intervals follow an ordinary law [are countable]), but the following demonstration appears to be more in the nature of things.
In the following passage, Borel let the closed, bounded interval be \(\left[A,B\right].\) He let \(\left(A_i,B_i\right)\) be one of the open intervals that contains \(A.\) If \(\left(A_i,B_i\right)\) doesn’t cover \(\left[A,B\right],\) then \(B_i\) is a point in the interior of \(\left[A,B\right]\) or is \(B\) itself. Let \(\left(A_{i_1},B_{i_1}\right)\) be one of the open intervals that contains \({B_i}.\) If \(B_{i_1}\) is less than or equal to \(B,\) then let \(\left(A_{i_2},B_{i_2}\right)\) be one of the intervals containing \(B_{i_1}.\) We continue in this fashion, stopping if we have covered \(\left[A,B\right].\)
Take the left endpoint \(A\) on the line, and let \(\left(A_i,B_i\right)\) be one of the intervals which contains the point \(A\); similarly let \(\left(A_{i_1},B_{i_1}\right)\) be one of the intervals which contains the point \(B_i\), let \(\left(A_{i_2},B_{i_2}\right)\) be one of the intervals which contains the point \(B_{i_1}\) etc. We suppose, as a matter of course, that \(A\) always designates the endpoint left of the intervals, \(B\) the endpoint right.
Notice that the process may terminate at this stage. Namely:
Case 1: If \(B_i\) is greater than \(B,\) or if \(B_{i_n}\) is greater than \(B\) for some \(n,\) then we have a finite cover of \(\left[A,B\right]\) given by \(\left(A_i,B_i\right)\) or by \(\left(A_i,B_i\right),\) \(\left(A_{i_1},B_{i_1}\right),\) \(\left(A_{i_2},B_{i_2}\right),\) ...,\(\left(A_{i_n},B_{i_n}\right),\) respectively. In Diagram 1, this would mean only the purple open set and a finite number of red intervals, including at least one purple or red interval with a right hand endpoint greater than \(B,\) are needed to cover \(\left[A,B\right].\)
Diagram 1: In our diagram of Borel’s argument, we see the closed interval \(\left[A,B\right],\) along with some of the open sets. Left endpoints are listed below the line and right endpoints and limit ordinals are above the line.
Otherwise, Borel examined the sequence of right endpoints \(B_{i_n}.\) This sequence is infinite (else we are in Case 1 above), bounded, and increasing, and thus converges. This is immediate by the monotone convergence property but can also be shown using a bit more work and the BolzanoWeierstrass property. Borel called the limit \(B_{i_{\omega}}\) (our first limit ordinal, above the green arrow in Diagram 1) and it is the supremum of the set \(\{B_{i_n}\}.\) Because the increasing sequence of right endpoints is bounded above by \(B,\) we know \(B_{i_{\omega}}\) is in \(\left[A,B\right].\) At this point, again the process may terminate:
Case 2: If \(B_{i_{\omega}}\) is equal to \(B,\) then we can form a finite cover in the following way. Obviously \(B\) and therefore \(B_{i_{\omega}}\) is contained in some open interval, which Borel called \(\left(A_{i_{\omega +1}},B_{i_{\omega+1}}\right).\) Since \(B_{i_{\omega}}\) is a limit, only a finite number of the \(B_{i_n}\) lie outside of \(\left(A_{i_{\omega +1}},B_{i_{\omega+1}}\right).\) Borel said that (for example) \(A_{i_{\omega +1}}\) lies between \(B_{i_{m1}}\) and \(B_{i_m}.\) Therefore, the finite number of intervals \(\left(A_i,B_i\right),\) \(\left(A_{i_1},B_{i_1}\right),\) \(\left(A_{i_2},B_{i_2}\right),\) …, \(\left(A_{i_{\omega}},B_{i_{\omega}}\right),\) \(\left(A_{i_{\omega +1}},B_{i_{\omega+1}}\right)\) cover the subinterval \(\left[A,B_{i_{\omega}}\right].\) In the case when \(B_{i_{\omega}}\) is \(B,\) this shows that a finite number of intervals covers the original interval. In Diagram 1, this would mean we need only one green open interval, one purple, and a finite number of red to cover \(\left[A,B\right].\)
The points \(B_{i_1},\) \(B_{i_2},\) \(B_{i_3},\) …, if they do not reach the extremity \(B\) of the line, have a limit \(B_{i_{\omega}}\) and this point is contained in an interval \(\left(A_{i_{\omega +1}},B_{i_{\omega+1}}\right)\) such that \(A_{i_{\omega +1}}\) falls, for example, between \(B_{i_{m1}}\) and \(B_{i_m};\) we will be able to disregard the intervals \(\left(A_{i_{m+k}},B_{i_{m+k}}\right),\) and we will have all the same an uninterrupted series of intervals on the line. We will continue similarly, passing to the limit when this is necessary and showing then that one can keep only a finite number of the already considered intervals.
If \(B_{i_{\omega}}\) and \(B_{i_{\omega +1}}\) are less than \(B,\) then we continue in this fashion. Borel already showed there was an infinite sequence of right endpoints \(B_i,\) \(B_{i_1},\) \(B_{i_2},\) … converging to \(B_{i_{\omega}}.\) By repeating the process starting with \(B_{i_{\omega}}\) (instead of \(B_i\)) we get an infinite sequence \(B_{i_{\omega+1}},\) \(B_{i_{\omega+2}},\) ... that converges to \(B_{i_{2\omega}}.\) This process could terminate here as well:
Case 3: If \(B_{i_{2\omega}}\) is equal to \(B,\) we can form a finite subcover of \(\left[A,B\right]\) in the following way. The passage above shows that we can cover \(\left[A,B_{i_{\omega}}\right]\) with a finite number of intervals. The argument in Case 2 then shows that we can cover \(\left[B_{i_{\omega}},B_{i_{2\omega}}\right]\) with a finite number of intervals. These two finite collections of intervals, taken together, cover \(\left[A,B\right].\) In Diagram 1, this means one purple, one blue, and a finite number of red and green intervals are needed to cover \(\left[A,B\right].\)
Assuming \(B_{i_{2\omega}}\) is less than \(B,\) Borel used the above process again, obtaining a sequence \(B_{i_{2\omega+1}},\) \(B_{i_{2\omega+2}},\) ..., converging to a limit \(B_{i_{3\omega}},\) which must be less than \(B\) (otherwise, we can use the technique of Cases 2 and 3 to get a finite cover). After exhausting all the \(B_{i_{m{\omega}+n}},\) he moved on to \(B_{i_{{\omega}^2}},\) ..., \(B_{i_{{{\omega}^2}+n}},\) ..., \(B_{i_{m{{\omega}^2}+n}},\) ..., \(B_{i_{{\omega}^p}},\) ..., \(B_{i_{{\omega}^{\omega}}},\) ..., all of which must be less than \(B\) (else we are done by the above analysis). This brings us to Case 4.
Case 4: Assume that all of the limit points are less than \(B.\) Borel stated that by looking at the right endpoints, he had created a surjection from the open cover to the sequence of right endpoints. Contained within the indices of the right endpoints are the following:
\[B_{i_1}, B_{i_2}, \dots,B_{i_{\omega}}, B_{i_{2\omega}}, \dots,B_{i_{{\omega}^2}}, \dots,B_{i_{{\omega}^{\omega}}}, \dots\]
which consists of all numbers of the second class (ordinals which have no immediate predecessor, or limit ordinals). Because the second class of numbers is not countable, and our original set was, he had arrived at a contradiction [17].
I say that we will necessarily reach the endpoint \(B\) of the line, because, if one did not reach it, one would define a series of intervals having for endpoints \[B_{i_1}, B_{i_2},\dots,B_{i_{\omega}}, B_{i_{2\omega}}, \dots,B_{i_{{\omega}^2}}, \dots,B_{i_{{\omega}^{\omega}}},\] the indices being all the numbers of the second class of numbers (defined by M. Cantor). But these indices are also in a certain order, the natural numbers, in entirety or in part. This is a contradiction because the second class of numbers constitutes a set of the second power.
One must be careful, because in actuality there need not be an interval with right endpoint of \(B_{i_{\omega}},\) \(B_{i_{2\omega}},\) \(B_{i_{{\omega}^2}},\) etc. It is not clear exactly what correspondence Borel referred to in this passage. On one hand, we can create a correspondence between limit ordinals and intervals in the following way
\[{B_{\alpha}}\leftrightarrow\left({A_{\alpha+1}},{B_{\alpha+1}}\right)\]
because these numbers do have successors. Or perhaps Borel used \(B_{i_{\omega}}\) to mean both the infinite sequence and the limit simultaneously. In either case, the flaw isn’t fatal, and he arrived at a contradiction for Case 4, thus completing the proof.
Here is an overview of Borel’s argument, which may be particularly useful to a teacher considering presenting this proof in class.
Background:

Completeness in the form of the monotone convergence property or perhaps the BolzanoWeierstrass property is needed to show that the right endpoints have a limit.

If presenting this proof, some cardinality issues must first be addressed. In particular, definitions of countable, ordinals, ordinal numbers of the second type, etc. will be necessary.
Benefits:

This is the first statement and proof of the HeineBorel Theorem and so has the benefit of priority.

We mentioned that there is a connection between the HeineBorel Theorem and completeness. As Hallet noted,
It was the original transfinite numbers proof which first established this connection. What the original proof shows, in effect, is that the BolzanoWeierstrass theorem employed a transfinite (but countable) number of times entails the HeineBorel property. [9, p. 23]

Koetsier and van Mill wrote that “Borel’s proof still counts as one of the first applications of transfinite ordinal numbers outside of set theory” [12]. Ironically, while students may feel that the HeineBorel Theorem is too abstract, it is an application of another abstract area of mathematics, namely Cantorian set theory.

In his 1898 restatement of his theorem, Borel mentioned that his proof was constructive and that it could be useful in actually creating the finite open cover [3, p. 42].
Passage 7. Borel noted that his proof was constructive.
Lebesgue in [13, p. 133] concurred, stating, “[I]t provides a regular procedure for forming the family, which is only logically defined by the other demonstrations which one gave.” Students who find the HeineBorel Theorem too abstract may appreciate that this technique gives an explicit covering.
Drawbacks:

The major drawback of the theorem is that it is not stated, or proved, in its full generality.

Providing a discussion of Cantor’s ordinal numbers may be a drawback if it is not naturally included elsewhere in the course.

Borel left out some of the finer points of the proof, which we have tried to remedy here. Having stated that “Borel actually gives very few of the details …” [9, pp. 2324], Hallet also expanded on the proof.
Impression:
We believe that if the HeineBorel Theorem is being taught during a set theory or measure theory course, this may be a particularly valuable proof to present. In addition, the constructive nature of the proof may benefit more computationally minded students.
Later Proofs:
Three years later, Borel published Leçons sur la théorie des fonctions [3]. Here he restated the theorem more carefully, noting that the covering must be countable in the statement of the theorem, rather than later in the proof. The proof he presented utilized the nested interval form of completeness and the “divide and conquer” technique that we will see in Cousin’s 1895 proof in the next section.
In May 1903, in “Sur l'approximation des nombres par des nombres rationnels” [4], Borel restated his theorem for arbitrary closed sets, not just intervals. He also mentioned that Lebesgue recognized that the covering need not be countable.
Define, in a space of n dimensions, a countable infinity of closed groups (that is to say such that each one contains its derivative) E_{1}, E_{2},…,E_{p},…, and an ordinary group E such that any point in E is interior to one of E_{i}. One can, therefore, choose among the E_{i} a limited number of groups such that any point of E is interior to one of them.
This theorem again, is not quite right – the set E needs to be closed and bounded. Certainly Borel used this fact in his proof, but Ernst Lindelöf (18701946) brought it to his attention in a letter on August 24, 1903 [8, p. 99]. When Borel published "Contribution à l’analyse arithmétique du continu" [2] later that year, he included the correct statement of the theorem.
Let E be a given closed and bounded group, E_{1}, E_{2},…,E_{p},…, a countable infinity of groups such that any point of E is interior to, at minimum, one of them; it is possible to find, among E_{1}, E_{2},…,E_{p},…, a limited number of groups such that any point of E is interior to, at minimum, one of them.