Intuition for Bolzano-Weierstrass Theorem First thing's first, let me write the down the definition of Bolzano-Weierstrass theorem that I'm using:

Bolzano-Weierstrass theorem for sequences: Every bounded sequence in $\mathbb{R}$ has a limit point in $\mathbb{R}$

My question is— How do I visualize this intuitively? Note that I'm not asking for a proof of this theorem, I only need some intuition behind it to internalise the idea.

What I have come across so far is that "Hah, since the sequence is bounded, that is each $|x_n|<M$ for some real $M$, for all $n$ and the fact that there are infinitely many terms in a sequence, the terms must huddle around a point".
I don't find this very convincing for instance, sure there are infinitely many terms in $(x_n)$ but the terms are only countably infinite (i.e. $n \mapsto x_n$) and the cardinality of $[-M, M]$ is greater than $\aleph _0$ so I don't see why each term $x_n$ can't be assigned to some real number $[-M,M]$ without huddling around a point. I'm confused.

If you can share any other intuition that'd be even more lovely. But I'd also like to know what is wrong with my counter-reasoning while you're at it.
 A: You are guaranteed to be able to pick out at least one monotonic (Non-increasing or non-decreasing) subsequence from your sequence.  All monotonic bounded sequences converge, therefore you have a subsequence converging to a point,  which will make it a limit point
A: Let $(a_n)_{n\ge 1}$ be a bounded sequence in $\mathbb{R}$. Without loss of generality, we assume that the sequence takes its values in $[0,1]$.
If the sequence contains any repeated value, then per definition that repeated value is a limit point. Therefore, if the sequence has no limit point, $\{a_n:n\ge 1\}$ is infinite. It suffices to show that every infinite subset $S\subseteq [0,1]$ has a limit point.
This assertion comes down to compactness of $[0,1]$. Assume that $S$ has no limit point, then for all $a\in [0,1]$ there exists some $\epsilon(a)>0$ such that the punctured open disk $\dot{U}_{\epsilon(a)}(a)$ contains no points of $S$.
Note that $\bigcup_{a\in [0,1]}U_{\epsilon(a)}(a)=[0,1]$, so by compactness there exists a finite subset $T\subseteq[0,1]$ with $\bigcup_{a\in T}U_{\epsilon(a)}(a)=[0,1]$. Now
$$S\subseteq[0,1]\setminus\bigcup_{a\in T}\dot{U}_{\epsilon(a)}(a) \subseteq T,$$
so $S$ is finite; a contradiction.
