Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits.

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 $\varepsilon > 0$ such that for each positive integer $N$, there exists some $j(N) > N$ 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.

• 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 $x'\ne x$.