TITLE: Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits.
QUESTION [17 upvotes]: Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits. 

My attempt is: 
Since the sequence is bounded , there exists $M>0$ such  that $x_n \in [-M,M]$ for all $n \in \mathbb{N}$. Since the sequence does not converge to $x$, there exists $\epsilon_0>0$ such that $ \forall N \in \mathbb{N}$, there exists $n \geq N$ such that $|x_n-x| \geq \epsilon_0$. 
Then we have $x_n \in [-M,x-\epsilon_0] \cup x_n \in [x+\epsilon_0,M]$. By Bolzano-weierstrass theorem, there exists a convergent subsequence in the two intervals. 
Is my proof valid? ${}{}$

REPLY [2 votes]: If a sequence $(x_n)$ is bounded, $a\le x_n \le b$ say, then it has at least one limit point $x$ with $a\le x\le b$ (Bolzano-Weierstraß) and 
a bounded sequence with exactly one limit point $x$ converges towards that limit point.

Therefore there must exist at least two distinct limit points and we can extract a converging sequence for each.
Just in case the second bullit point above is not clear: 
If $a\le x_n\le b$ for all $n$ and $x$ is not the limit of $x_n$, then  there exists $\epsilon>0$ such that infinitely many $x_n$ are outside $(x-\epsilon,x+\epsilon)$, hence there are infinitely many $x_n>x+\epsilon$ or infinitely many $x_n<x-\epsilon$, leading to at least one limit point $x'$ in $[x+\epsilon,b]$ or $[a,x-\epsilon]$. Thus if $x$ is a limit point but not the limit, there is another limit point $x'\ne x$.