Lemma 2.6.3 $\implies (x_{n})$ is bounded. So use the Bolzano-Weierstrass Theorem to produce a convergent subsequence $(x_{n_{k}})$ . Set $x= \lim x_{n_{k}}.$ So $(x_{{n_{k}}}) \to x. \quad \Large{(♪)}$. The idea is to show that the original sequence $(x_{n})$ converges to this same limit. Once again, use a triangle inequality argument. $\color{brown}{\text{ We know the terms in the subsequence are getting close to the limit $x$ } }$. $\color{forestgreen}{\text{ $(x_{n})$ is assumed to be Cauchy, so the terms in the tail of $(x_{n})$ are near each other. } \qquad (♫)}$ Thus, make each of these distances less than half of the prescribed $\epsilon$.
1. Where are the 'ideas' of this proof on top from? I cannot predict things, and I don't want to memorize all these steps to remember how to prove this on the exam. Is there any intuition?
Let $\epsilon >0$. Because $(x_{n})$ is Cauchy, there exists $N$ such that $|x_{n}-x_{m}|< e/2$ whenever $m,\ n\geq N$.
By $(♪)$, choose a term in this subsequence, call it $x_{n_{m}}$, with $ n_{m}\geq \color{red}{N}$and $ |x_{n_{m}}-x|<\frac{\epsilon}{2}.$
2. Why $ n_{m}\geq \color{red}{N}$?
$(♪)$ implies that for all $\varepsilon/2 > 0$, there exists $N_1$ such that $n_m \ge N_1 \implies |x_{\Large{n_{m}}} -x|< \varepsilon/2$.
But $N_1$ doesn't relate to $\color{Red}{N}$? Why didn't proof write what I wrote in this box?
$n_{m}$ has the desired property for the original sequence $(x_{n})$, because if $n \ge n_{m}$. then $\begin{align} |x_{n}-x|\ & =\ |x_{n}-x_{\Large{n_{m}}} \; + x_{\Large{n_{m}}} - x| \\ & \leq\ \color{forestgreen}{|x_{n}-x_{\Large{n_{m}}}|}+ \color{brown}{|x_{\Large{n_{m}}}-x|} \\ & <\frac{\epsilon}{2}+\frac{\epsilon}{2} \end{align}$
3. How does the green sentence $\color{forestgreen}{(♫)}$ signify $\color{forestgreen} { |x_{n}-x_{\Large{n_{m}}}| }$ ?
I think gowers.wordpress.com expatiates on this here.
For 1, the intuition really comes from just being comfortable with what the definitions are really saying. The proof can be summed up fairly nicely as
So the terms of $(x_n)$ should be tending to the limit of the $x_{n_k}$ as well. As you've written, the definitions make this all precise, but it's really just a matter of reflecting on what Cauchy/convergent mean: points getting close to each other/a limit.
For 2, it's perhaps a little sloppy to do this if they haven't gone over this before, but you are correct that the $N$ and the $N_1$ do not have to be related. However, the condition is for all $n \geq N$ and for all $n_m\geq N_1$, so set $N' = \max(N,N_1)$ and you can use the same $N'$ for both the original sequence being Cauchy and the subsequence converging.
Why? Any $n\geq N'$ is certainly greater than $N$ by the choice of $N'$, and similarly any $n_m\geq N'$ must be greater than or equal to $N_1$, so the conditions we want hold.
They seem to have glossed this here and just set $N$ to be something large enough, like the $N'$ chosen above. You can always pick a larger $N$ because the condition is for all $n$ larger anyway, so this is a detail that will likely be glossed over for the rest of the book and is good to be comfortable with!
For 3, $(x_n)$ is Cauchy. So, you know that for $n,m\geq N$ we have $|x_n - x_m| < \epsilon/2$. So if $n,n_m\geq N$, the last line follows. But we chose them so that $n \geq n_m \geq N$, and everything works out!