Using the *compact* property of subsets in $R$ to prove Bolzano-Weierstrass Theorem

Question

I have been asked to prove the Bolzano-Weierstrass Theorem with respect to a bounded sequence of real numbers by using the fact that closed and bounded subsets of $R$ are compact.

There is a hint along with the question stating that there are two cases: one where the same value appears infinitely many times in the sequence, and another where each value appears a finite number of times.

I have been attempting this question for a while, but I have been having trouble relating the compact property to the proof.

Answer 1

Hint 1 : Every sequence $a_n$ contains a monotone subsequence.

The proof is hidden below, you can attempt this yourself or look at the proof (by hovering the mouse over it) if you like:

Let us call $n \in \mathbb{N}$ special if $a_n > a_m$ for all $m> n$. Then, two possibilities arise. One is that there are infinitely many special points, in which case there is a decreasing subsequence. On the other hand, if there are finitely many special points, then take $n_1$ larger than all the special points. Since $n_1$ is not special, there is $n_2 > n_1$ such that $a_{n_2} \geq a_{n_1}$. Now $n_2$ is not special etc. so we can form an increasing subsequence.

Hint 2:A bounded monotone sequence converges. This proof I leave to you.

Hence, every bounded sequence has a monotone subsequence, which is also bounded hence converges. You don't at all need compactness in this proof.