Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits. 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 $M&gt;0$ such  that $x_n \in [-M,M]$ for all $n \in \mathbb{N}$. Since the sequence does not converge to $x$, there exists $\epsilon_0&gt;0$ 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? ${}{}$
Use Bolzano-Weierstrass to extract a subsequence $x_{i_1}, x_{i_2}, \dotsc$ that converges to some $a$. Since $x_1, x_2, \dotsc$ does not converge to $a$, there exists some $\varepsilon &gt; 0$ such that for each positive integer $N$, there exists some $j(N) &gt; N$ such that $|x_{j(N)} - a| \geq \varepsilon$. Use Bolzano-Weierstrass to extract from $x_{j(1)}, x_{j^2(1)}, x_{j^3(1)}, \dotsc$ a subsequence $x_{k_1}, x_{k_2}, \dotsc$ that converges to some $b$. Clearly, $a \neq b$.
By the way, this method works for any $x_1, x_2, \dotsc$ in $\mathbb{R}^n$, which goes beyond what the OP intended.
Yes, valid, but could write it even clearer.

For each $N$ there is an $n$ such that $|x_n-x|\ge \epsilon_0$. Let such an $n$ be chosen for all $N$ and denote it by $k(N)$, for example. Thus we get a subsequence $(x_{k(1)},x_{k(2)},\dots)$ of $(x_n)$, which lies in $[-M,x-\epsilon_0]\cup [x+\epsilon_0,M]$, so this subsequence has a convergent subsequence, and the limit of that cannot be $x$.