Here's the statement:
A sequence $(x_n)$ is bounded iff every subsequence of $(x_n)$ has a convergent subsequence.
Okay so the $(\Rightarrow) $ part is easy to prove. Since $(x_n)$ is bounded, any of its subsequences are bounded thus by the Bolzano Weierstrass Theorem, each of those subsequences have a convergent subsequence.
Now, the fate of the statement lies on the $(\Leftarrow ) $ part. However, I neither could prove it nor provide a counterexample. Hints will be appreciated!
Here's what I tried though: Since a subsequence of a subsequence of a sequence is itself a subsequence of that sequence. Hence, we know there are some convergent subsequences of that particular sequence! I'm not sure how this helps.
If the sequence is unbounded, then there is a $n_1\in\mathbb N$ such that $|x_{n_1}|>1$. And there is a $n_2>n_1$ such that $|x_{n_2}|>2$. And so on. Obviously, $\left(x_{n_k}\right)_{k\in\mathbb N}$ doesn't converge.