There are a variety of methods for proving the Bolzano-Weierstrass theorem.
One (such as the one given on wikipedia) relies on extracting a monotone subsequence.
Another, like the one you are reading I suspect, relies on the nested interval theorem: Given any nested sequence of closed bounded intervals $I_0 \supset I_1 \supset I_2 \supset \cdots$ there exists a point in the intersection $\cap_{k=0}^\infty I_k$.
A sketch of the rest of the proof: take a point (by the nested interval theorem) in the intersection of all the intervals you're considering. Then show that this point is in fact the limit of the subsequence, since the length of the intervals is going to zero.
I'll give some more details. The proof is as follows:
Let $M$ be an upper bound on the absolute values of the terms in your sequence. Let $I_0 = [-M,M]$. Let $a_0$ be the first term in your sequence. Note that infinitely many terms in the sequence are in $I_0$ (in fact they all are).
We inductively construct a subsequence. Given $I_k$ such that infinitely many terms in the sequence are in $I_k$, let $I_{k+1}$ be the closed left half of $I_k$ if there are infinitely many terms in the sequence in the closed left half. Otherwise, let $I_{k+1}$ be the closed right half of $I_k$ (necessarily with infinitely many terms). Let $a_{k+1}$ be the smallest term in your sequence which is both in $I_{k+1}$ and not already used.
Lemma: given $x \in I_k$, we have, for $j \geq k$ that $|x - a_j| \leq \frac{M}{2^k}$
Let $y \in \cap_{k=0}^\infty I_k$, which exists by the nested interval theorem.
For $j \geq k$, we have $|y - a_j| \leq \frac{M}{2^k}$ by the lemma, since $y \in I_k$. Thus the subsequence we have constructed converges to $y$.
One proof of the nested interval theorem: The left endpoints of the intervals form a monotone sequence. The limit of this sequence is in every one of the intervals.
While it's a slightly longer proof than the monotone subsequence proof, I find it highly intuitive, and the nice thing about this proof is that it generalizes well to higher dimensions.
If $X$ is a normed vector space then we have:
$ \dim X < \infty \iff X$ has the Bolzano- Weierstraß- property.
$X$ has the Bolzano- Weierstraß- property, if every bounded sequence in $X$ contains a convergent subsequence.
Now let $X= \mathbb R^{n\times n}$ be equipped with any norm. Convergence in this norm is element wise convergence.
Can you take it from here ?
Best Answer
The Arzela-Ascoli theorem states: consider a family of real-valued functions $\mathcal{F} = \{f_n \vert n \in \mathbb{N}\}$ defined on a compact set. If $\mathcal{F}$ is uniformly bounded and equicontinuous, then it is precompact.
To answer your question, if a given sequence $(f_n)$ meets the boundedness and equicontinuity conditions, then there exists a uniformly convergent subsequence $(f_{n_k})$.
There are other, equivalent statements of the theorem, but I believe this will be the most useful for your purposes.