To prove the Bolzano-Weierstrass theorem, we will first need two lemmas.

Lemma 1.

All bounded monotone sequences converge.

proof.

Let $(s_n)$ be a bounded, nondecreasing sequence. Let $S$ denote the set $\{s_n : n \in \mathbb{N}\}$ Then let $b=\sup S$ (the supremum of $S$ )

Choose some $\epsilon > 0$ Then there is a corresponding $N$ such that $s_N>b-\epsilon$ Since $(s_n)$ is nondecreasing, for all $n>N$ $s_n > b-\epsilon$ But $(s_n)$ is bounded, so we have $b-\epsilon < s_n \le b$ But this implies $|s_n-b|<\epsilon$ so $\lim s_n= b$ $\square$ (The proof for nonincreasing sequences is analogous.)

Lemma 2.

Every sequence has a monotonic subsequence.

proof.

First a definition: call the $n$ term of a sequence dominant if it is greater than every term following it.

For the proof, note that a sequence $(s_n)$ may have finitely many or infinitely many dominant terms.

First we suppose that $(s_n)$ has infinitely many dominant terms. Form a subsequence $(s_{n_k})$ solely of dominant terms of $(s_n)$ Then $s_{n_{k+1}} < s_{n_k}$ $k$ by definition of ``dominant'', hence $(s_{n_k})$ is a decreasing (monotone) subsequence of ($s_n$ .

For the second case, assume that our sequence $(s_n)$ has only finitely many dominant terms. Select $n_1$ such that $n_1$ is beyond the last dominant term. But since $n_1$ is not dominant, there must be some $m>n_1$ such that $s_m > s_{n_1}$ Select this $m$ and call it $n_2$ However, $n_2$ is still not dominant, so there must be an $n_3>n_2$ with $s_{n_3} > s_{n_2}$ and so on, inductively. The resulting sequence

$$ s_1,s_2,s_3,\ldots $$

is monotonic (nondecreasing). $\square$ proof of Bolzano-Weierstrass.

The proof of the Bolzano-Weierstrass theorem is now simple: let $(s_n)$ be a bounded sequence. By Lemma 2 it has a monotonic subsequence. By Lemma 1, the subsequence converges. $\square$