Why do proofs using continuity use the Bolzano-Weierstrass theorem?

Question

This may just be goofy of me but there's a lot of proofs I've been seeing recently while learning real analysis that keep on invoking the Bolzano-Weierstrass theorem, and while I understand how it's being used and it makes sense, I don't quite understand why it needs to be used at all.

For example, this was a proof of the extreme value theorem,

If $f$ is continuous, then $f$ is bounded by the previous theorem. Thus, the set $E = \{f(x) | x \in [a, b]\}$ is bounded above. Let $L = \sup E$. Then,

1. $L$ is an upper bound for $E$, i.e. $\forall x \in [a, b], f(x) \le L$.
2. There exists a sequence $\{f(x_n)\}$ with $x_n \in [a, b]$ such that $f(x_n) \to L$.
   By the Bolzano-Weierstrass theorem, there exists a subsequence $\{x_{n_k}\}$ of ${x_n}$ and $d \in [a, b]$ such that $x_{n_k} \to d$ as $k \to \infty$. Hence, $f(d) = \lim_{k \to \infty} f(x_{n_k}) = \lim{n \to \infty} f(x_n) = L$ by the continuity of $f$. Thus, $f$ achieves an absolute maximum at $d$.

I understanding creating a sequence in $[a,b]$ such that $f(x_n)$ approaches $L$. I also understand that for the sequence you can generate a subsequence such that $x_{n_k}$ approaches $d$. I'm just a little lost on why we can't just say that ${x_n}$ approaches $d$, and why we need to create a whole subsequence just to justify it. If there's anyone who can help me out here, many thanks.

Answer 1

Here's an example: let $f(x) = |x|$ on $[-1,1]$. The sequence $x_n = (-1)^n(1-1/n)$ satisfies $f(x_n) \to 1$, but the sequence is not convergent. Nothing in the proof rules out $x_n$ being a sequence like that, which is why you need to take a subsequence.

Answer 2

The underlying reason is simply beacuse the B-W is a strong result. In general, we know very little about sequnces, but a lot about converging sequances, so the B-W theorem provides some insight about general sequances. In the proof you described, the problem is that we do not know a lot about $x_n$. we cannot concluse that $x_n->d$. for example, the sequance $x_n = (-1)^n$ is clearly not converging but the subsequance $x_{2n}$ is constant, thus converges to 1. In a broader sense, the B-W theorem shows that a closed interval in $\Bbb R$, namely $[a, b]$ with $a<b$ is a compact set (which you will find later that it is a really useful trait).