Two subsequences with different limits limplies **diverge** Problem_

Prove that if two subsequences of a sequence $\{a_n\}$ have different limits $a \ne b$, then $\{a_n\}$ diverges.

In fact, I've seen several proofs of the question. But, unfortunately, I cannot understand some of those solutions. One proof that I saw the most is:

Pick a value of $c$. We wish to show that the sequence does not converge to $c$. Now, either $b≠c$ or $a≠c$ (or both). (1) By WLOG, assume $a≠c$. 
Then let $ϵ=\frac{|a−c|}{2}$. Now let N be any natural number.
Our task, now, is to find some $s>N$ so that $|c−x_s|\geϵ$. Since $\{x_{n_i}\}$ converges to $a$, (2) let $s=n_i$, where $n_i$ is large enough so that $|a−x_{n_i}|<ϵ$ and $n_i\gt N$. 
Then
  $$|c−a|=|c−x_s+x_s−a|≤|c−x_s|+|a−x_s|<|c−x_s|+ϵ=|c−x_s|+\frac{|c−a|}{2}$$
  Therefore,
  $|c−x_s|>\frac{|c−a|}{2}=ϵ$
(3) And we're done.

See more at the following site:
[1] Two subsequences with different limits $\implies$ not convergent
[2] If a sequence has two convergent subsequences with different limits, then it does not converge
Here are my questions:
Questions_
(1) How can 'WLOG' be applied in this case?
(2) Can we set $s$ as $n_i$ without any other conditions?
(3) The solution only analyzed the subsequence converges to $a$. How about the other one(converges to $b$)? Is it covered by WLOG? Then, how?
Thanks for giving me ideas.
 A: *

*"WLOG" means "without loss of generality." In this case, we know either $b \neq c$ or $a \neq c$. But there really is no distinction between $a$ and $b$ in this problem - they are both limits of some subsequence of $\{a_n\}$, but there is no other distinction between them. Thus we could safely "rename" them if we wanted to in order to have $a \neq c$. An alternative way of looking at it is that we could go through the proof under the assumption $a \neq c$, then go through the analogous proof again under the assumption that $b \neq c$, and come to the same conclusion in either case. Saying "WLOG, $a \neq c$" allows us to avoid writing essentially the same proof twice for the two different cases.

*I'm not sure exactly what you're asking here. But we know that since a subsequence of $\{a_n\}$ converges to $a$, there must be some $M$ such that for all $n_i>M$, $|a_{n_i}-a|<\epsilon$, where $n_i$ is the sequence of indices for the subsequence converging to $a$. Now we want $n_i>N$, and since we know the above holds for all $n_i>M$, we can certainly find an $n_i>N$ for which it holds. If you'd like, take $n_i$ to be the smallest $n_i$ greater than $\max(M,N)$.

*See point 1. 

A: 1.) WLOG stands for Without Loss Of Generality. In this case it is being used to signify that although in our assumption either ≠ or ≠, it doesn't matter which is true so for the proof we will just assume that ≠ to make the writing cleaner.
2.) Here we are setting $s$ as an index for for our sequence $a_i$, and the only conditions we are putting on it is that is that for all indices greater than $n_i$, $x_{n_i}$ is in the epsilon neighborhood of a.
3.) Yes we are done at this point. We can apply the WLOG here because in the case where $a\neq c$ we've already done the work, but if $b \neq c$ then our argument is any different, we just replace every instance of $a$ with $b$ and the argument still holds, so it suffices to prove the case where $a \neq c$. (And in the case where $a \neq c$ and $b \neq c$ then we can just use the fact that $a \neq c$ and ignore the other part to prove it exactly in the way you cited.)
