Theorem. Let $I:=[a,b]$ be a closed bounded interval and let $f:I\to \mathbb{R}$ be continuous on $I$. Then $f$ has an absolute maximum and absolute minimum on I.
Note that the absolute maximum portion is all I need to answer my question in this proof.
Proof.
By hypothesis the set $f(I) := \{f(x): x\in I\}$ is bounded in $\mathbb{R}$ [proven in previous thm]. Let $s = \sup f(I)$. I claim that that there exists an $x_s$ in $I$ such that $s = f(x_s)$.
Since $ s = \sup f(I)$, if $n \in \mathbb{N}$, then the number $s - 1/n$ is not an upperbound of the set $f(I)$. Consequently there exists a number $x_n$ s.t.
$$s - 1/n < f(x_n) \leq s \, , \, \, \forall n \in \mathbb{N} \tag{$\star$}$$
Since $I$ is bounded, the sequence $X = (x_n)$ is bounded. Therefore, by the Bolzano-Weierstrass Theorem, there exists a subsequence $X' = (x_{n_k})$ of $X$ that converges to some number $x_s$. Since the elements of$X'$ belong to $I$ it follows that $x_s\in I$. Therefore $f$ is continuous at $x_s$ so that $\lim(f(x_{n_k})) = f(x_s)$. Applying this fact to ($\star$) and using the squeeze theorem we conclude that $\lim(f(x_{n_k})) = s$ and we have
$$f(x_s) = \lim(f(x_{n_k})) = s = \sup f(I)$$
which means $x_s$ is an absolute maximum point of $f$ on $I$.
The part I'm having difficulty with is the the "$\leq$" in inequality $(\star)$. Why can we assume that $f(x_n)$ might be equal to $\sup f(I)$? It seems as though we're assuming that which is to be proved might be true.
And furthermore, I can not recall using the squeeze theorem on strict inequalities (the left side of $(\star)$.
We are considering the possibility that $f(x_n)=s$, we are not assuming this is the case. It would be wrong to say $f(x_n)<s$ (the theorem shows this is not always the case!).
To put this another way, we do not know that the equality is obtained (you are correct, this is what we are trying to prove). However, by definition of the supremum, we can say with certainty that for any $x\in I$ that $f(x)\le s$.