How and why do we prove that $\mathcal{l}(I) \leq \sum_{i=1}^\infty \mathcal{l}(J_i)$ where $I=\bigcup_{i=1}^\infty J_i$ is an open interval? Leading up to a treatment on Lebesgue outer measure, I have been shown a proof that

$\mathcal{l}(I) \leq \sum_{i=1}^\infty \mathcal{l}(J_i)$ for $I=\bigcup_{i=1}^\infty J_i$ is an open interval (where $\mathcal{l}$ is the length function, i.e. $\mathcal{l}(I)=b-a$ if $I=(a,b)$).

The proof creates slightly larger intervals for each $J_i=(a_i,b_i)$ by adding an amount $\epsilon>0$ to each endpoint and considers a finite open cover of the closed interval $[a,b]$ by these enlarged $J_i$'s.
My question is, what is the purpose of this discussion, what does it show us, and what is the intuition behind the proof/why does this need to be proved at all?
 A: Since we started discussing this particular proof in the comments, I will copy here the proof from Oxtoby's book Measure and Category. (The second book in the series Graduate Texts in Mathematics.)

Theorem 1.5 (Borel). *If a finite or infinite sequence of intervals $I_n$ covers an interval $I$, then $\sum|I_n|\ge|I|$.
Proof. Assume first that $I=[a,b]$ is closed and that all of the intervals $I_n$ are open. 
  Let $(a_1,b_)$ be the first interval that contains $a$. 
  If $b_1\le b$, let $(a_2,b_2)$ be the first interval of the sequence that contains $b_1$.
  If $b_{n-1}\le b$, let $(a_n,b_)$ be the first interval that contains $b_{n-1}$
  This procedure must terminate with some $b_N>b$. 
  Otherwise the increasing sequence $\{b_n\}$ would converge to a limit $x\le b$ and $x$ would belong to $I_k$ for some $k$. 
  All but finite number of intervals would have to precede $I_k$. 
  This is impossible since no two of these intervals are equal.
  (Incidental1y, this reasoning reproduces Borel's own proof of the "Heine-Borel theorem" [5, p.228].)
  We have
  $$b-a < b_N-a_1 = \sum_{i=2}^N (b_i-b_{i-1}) +b_1-a_1 \le \sum_{i=1}^N (b_i-a_i),$$
  and so theorem is true in this case.
In the general case, for any $\alpha>1$ let $J$ be a closed subinterval of $I$ with $|J|=|I|/\alpha$, and let $J_n$ be an open interval containing $I_n$ with $|J_n|=\alpha|I_n|$.
  Then $J$ is covered by the sequence $\{J_n\}$.
  We have already shown that $\sum|J_n|\ge|J|$.
  Hence $\alpha\sum|I_n| = \sum|J_n| \ge |J| \ge |I|/\alpha$. Letting $\alpha\to1$ we obtain the desired conclusion.
[5] Borel, E.: Lecons sur la theorie des fonctions. Paris: Gauthier-Villars 1914.


The OP asked about this particular step:

All but finite number of intervals would have to precede $I_k$. 

If we have that $b_n\to x$ and $x$ belongs to an open interval $I_k$, then starting from some $n$ each $b_n$ belongs to $I_k$. (From the definition of limit.) But since $(a_{n+1},b_{n+1})$ was the first interval from the sequence $I_1,I_2,\dots,I_k,\dots$ which contains $b_n$, the interval $I_k$ must come in this sequence at some point after the interval $(a_{n+1},b_{n+1})$. This leads to a contradiction, since $I_k$ is preceded only by finitely many intervals (since $k$ is some finite number) and, at the same time, in the proof we constructed infinitely many intervals of the form $(a_n,b_n)$ which precede $I_k$.
