A while back I posted a long proof of the Bolzano-Weierstrass theorem -- also known as the "sequential compactness theorem" -- which basically says every sequence that's bounded has a subsequence within it that converges. Here's a much shorter and simpler version of it.
First we'll prove a lemma that shows for any sequence we can always find a monotone subsequence -- that is, a subsequence that's always increasing or decreasing.
Lemma. Every sequence has a monotone subsequence.
Proof. Let be a sequence. Define a "peak" of as an element such that for all . That is is a peak if forever after that point going forward, there is no other element of the sequence that is greater than . Intuitively, think of shining a flashlight from the right onto the "mountain range" of a sequence's plotted elements. If the light hits an element, that element is a peak.
If has infinitely many peaks, then collect those peaks into a subsequence . This is a monotone decreasing subsequence, as required.
If has finitely many peaks, take n to be the position of the last peak. Then we know is not a peak. So there exists an such that . Call this point . We also know that is not a peak either. So there also exists an such that . Call this point .
Continuing, we can create a subsequence that is monotone increasing. In either case -- if our sequence has infinite or finitely many peaks -- we can always find a monotone subsequence, as required.
Now that we've proved the above lemma, the proof of the Bolzano-Weierstrass theorem follows easily.
Theorem (Bolzano-Weierstrass).Every bounded sequence has a convergent subsequence.
Proof. By the previous lemma, every sequence has a monotone subsequence. Call this . Since is bounded by assumption, then the subsequence is also bounded. So by the monotone convergence theorem, since is monotone and bounded, it converges. So every bounded sequence has a convergent subsequence, completing the proof.