Proof explanation of Borel Theorem on $\mathbb{R}$ Notation: The length of any interval $I$ is denoted by $|I|.$
In the book Measure and Category by Oxtoby, in Chapter $1$ Measure and Category on real line, he states the following theorem and proof.

Theorem $1.5$ (Borel) If a finite or infinite sequence of intervals $I_n$ covers an interval $I,$ then $\sum|I_n| \geq |I|.$
Proof: Assume first that $I=[a,b]$ is closed and that all of the intervals $I_n$ are open. 
  Let $(a_1,b_1)$ be the first interval that contains $a.$
  If $b_1\leq b,$ let $(a_2,b_2)$ be the first interval of the sequence that contains $b_1.$
  If $b_{n-1}\leq b,$ let $(a_n,b_n)$ ne 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 $x\leq b,$ and $x$ would belong to $I_k$ for some $k.$
  All but a finite number of the intervals $(a_n,b_n)$ would have to precede $I_k$ in the given sequence, namely, all those for which $b_{n-1}\in I_k.$
  This is impossible, since no two of these intervals are equal.

I have two questions: 
$(1)$ What is the meaning of 'first interval' in the proof? 
For example, let $[a,b]=[0,1].$ The two intervals $(-1,0.5)$ and $(-0.1, 1)$ contain $a=0.$ Which interval should we choose for $(a_2,b_2)?$
$(2)$ I do not understand how author conclude the bold statement in the proof.
My understanding is that since $x$ is the limit of $\{b_n\},$ then for any open interval $I_k$ containing $x,$ by definition of limit, the 'tail' of the sequence $\{b_n\}$ is contained in $I_k.$
However, I fail to 'see' any contradiction in this situation. 
Any help would be appreciated. 
 A: Oxtoby is a lovely author, but I just learned he probably did not read Borel's proof because I have just read his actual proof, and Q2 has a horrible answer.
First Q. First means, from left to right. So in your example, $b_1=0.5$ and $b_2=1$. But remember you have to cover $[0,1]$. Your example is not a covering since $1\notin (-1,0.5)$ and $1\notin(-0.1,1)$.
Second Q. Let me just write a better proof for everyone who comes through here (only the compact case though).
Let $I=[a,b]$. Suppose we have a cover $\{I_n\}$ of $I$.
Simplest Route (uses Heine-Borel): Since $I$ is compact, $\{I_n\}$ has a finite sub-cover $\{I_{j_i}\}_{i=1}^n$. Since $I\subseteq \cup I_{j_i}$, this implies
$$
|I|\leq \sum_{i=1}^n |I_{j_i}|\leq\sum_{i=1}^\infty |I_{i}|
$$
so we are done.
Oxtoby/Borel (does not use Heine-Borel):
Read this.
The link shows the complete proof. Essentially, Oxtoby is trying to show a finite sub-cover of $I$, but the way he goes about it is not-at-all obvious. More precisely, it seems Oxtoby has made some kind of assumption using a priori knowledge. 
