proof understanding : $\lim x_{n_k} = \limsup x_n$ 

It is a bit hard for me to understand the proof. 


*

*In the first yellow line, does this mean that we make a subsequence $\{x_{n_1},x_{n_2},....,x_{n_k}\}$ = $\{x_1,x_2,....x_k\}$ ??

*I don't understand why $a_{(n_k+1)}=$sup $\{x_n:n\ge n_k+1\}$
Thank you in advance!
 A: Point 1: No, it doesn't -- it just means that $n_1 = 1$. We could have $n_2 = 30$, and $n_3 = 25295$, for example. We just have to pick the $n_k$ so that $x_{n_{k+1}}$ gets very close to $a_{n_k+1}$.
Point 2: the author's notation is a bit annoying here. In the first line of the proof, $a_n$ is defined as
$$
a_n = \sup\{x_k:k \ge n\}
$$
Now swap the symbol "$n$" with "$n_k + 1$" and the symbol "$k$" with "$n$". This gives us
$$
a_{n_k+1} = \sup\{x_n:n\ge n_k+1\}
$$
and for some reason the author puts parentheses on the subscript of $a_{n_k+1}$... Clearly they're equivalent, though. In other words, we're just letting $n := n_k+1$ and using the name "$n$" instead of "$k$" in the expression above so that we aren't confusing it with the $k$ in $n_k$ (which is something completely different).
A: Here is a different proof. 
Note that there exists a subsequence of $(x_n)_n$ that converges to $x$ iff  whenever $J$ is an open interval containing $x,$ the set $\{n: x_n\in J\}$ is infinite.
Suppose by contradiction that no subsequence of $(x_n)_n$ converges to $x.$ Then let $r,s>0$ such that $\{n: x_n\in (-s+x,r+x)\}$ is finite.
Now the set $\{n:x_n\geq r+x\}$ is also finite. Otherwise $a_m=\sup_{n\geq m}x_n> r+x  $ for  all  $m,$ implying $x=\lim_{m\to \infty}a_m\geq r+x>x,$ which is absurd.
So  $x_n\leq -s+x$ for all but finitely many $x.$ This implies that for some $M\in \Bbb N$ we have $n\geq M\implies x_n\leq-s+x,$ and hence we have $n\geq M\implies a_n\leq -s+x,$ but this  implies $x=\lim_{n\to \infty}a_n\leq -s+x<x, $ which is absurd.
