= Solution
The <Bolzano-Weierstrass theorem> states that every bounded sequence of real numbers has a convergent subsequence.
For a proof, place all terms in a closed bounded interval $I_1$. Bisect it and choose a closed half $I_2$ containing infinitely many terms. Continue inductively, choosing nested closed intervals
$$
I_1\supseteq I_2\supseteq\cdots
$$
with infinitely many sequence terms and with lengths tending to zero. Choose $n_k>n_{k-1}$ such that $x_{n_k}\in I_k$. The <nested interval theorem> gives a unique point $x$ in every $I_k$. Because both $x_{n_k}$ and $x$ lie in $I_k$,
$$
|x_{n_k}-x|\leq |I_k|\longrightarrow0.
$$
Thus $(x_{n_k})$ is a convergent subsequence.
Now suppose every convergent subsequence of the bounded sequence $(x_n)$ converges to $L$. If $x_n$ did not converge to $L$, there would be an $\epsilon>0$ and a subsequence $(x_{n_k})$ satisfying
$$
|x_{n_k}-L|\geq\epsilon
$$
for every $k$. This subsequence is bounded, so Bolzano--Weierstrass gives a convergent subsubsequence. By hypothesis its limit is $L$, contradicting the displayed inequality. Hence the <unique subsequential limit of a bounded sequence> principle gives
$$
\boxed{x_n\to L}.
$$
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