Here is a proof of the Extreme Value Theorem that does not need to extract convergent subsequences. First we prove that :
Lemma: Let $f : [a,b] \rightarrow \mathbb{R}$ be a continuous function, then $f$ is bounded.
Proof: We prove it by contradiction. Suppose for example that $f$ does not have an upper bound, then $\forall n\in\mathbb{N}$, the set $\{x \in [a,b] , \, f(x) \geqslant n\}$ is not empty. Consider the following quantity:
$$a_n = \inf\{x \in [a,b] , \, f(x) \geqslant n\}.$$
For all $n\in \mathbb{N}$, $a_n \in [a,b]$ exists. By the continuity of $f$, $f(a_n) \geqslant n$. And since $\{x \in [a,b] , \, f(x) \geqslant n+1\} \subset \{x \in [a,b] , \, f(x) \geqslant n\}$, we have $a_{n+1} \geqslant a_n$. Since $(a_n)_{n\in\mathbb{N}}$ is a monotonic bounded sequence, it has a limit:
$$a_{\infty} = \lim_{n\to\infty} a_n $$
and $a_{\infty} \in [a,b]$. Let $M = \lceil f(a_{\infty}) \rceil$, then $\forall n \geqslant M+2$, $f(a_n) > f(a_{\infty})+1$. Hence by the continuity of $f$,
$$f(a_{\infty}) = \lim_{n\to\infty}f(a_n) \geqslant f(a_{\infty})+1,$$
which yields a contradiction. Therefor $f$ must have an upper bound on $[a,b]$. For the same reason, $f$ must have an lower bound on $[a,b]$. In conclusion, $f$ is bounded on $[a,b$
With this lemma we can prove the Extreme Value Theorem.
Theorem: Let $f : [a,b] \rightarrow \mathbb{R}$ be a continuous function, then $f$ has at least a maximum and a minimum point.
Proof: We proved in the lemma that $f$ is bounded, hence, by the Dedekind-completeness of the real numbers, the least upper bound (supremum) $M$ of $f$ exists. By the definition of $M$,
$$\forall n \in \mathbb{N}, \, S_n = \{x \in [a,b] , \, f(x) \geqslant M - \frac1n\} \neq \emptyset .$$
Let $s_n = \inf S_n$ be the infimum of $S_n$. We know that $a \leqslant s_n \leqslant s_{n+1} \leqslant b$ and $f(s_n) \geqslant M - \frac1n$. Since $(s_n)_{n\in\mathbb{N}}$ is a monotonic bounded sequence, the limit $s = \lim_{n\to\infty} s_n \in [a,b]$ exists.
$\forall N\in\mathbb{N}$, $\forall n > N$, $f(s_n) > M - \frac1N$, so $M \geqslant f(s) = \lim_{n\to\infty} f(s_n) \geqslant M - \frac1N$. Hence $f(s) = M$ and $f$ has at least a maximum point. Similarly we can prove that $f$ has at least a minimum point. Q.E.D
You are basically done. Here are 2 proofs:
WLOG assume f is not a constant. For every ε we can find an α such that for all x outside of (-α,α), |f(x)| is ε-small.
Since f is not constant it has a supremum S and infimum I over the whole space with S≠I. Since f is bounded, S and I are not infinite. Since S≠I, at least one is not zero. Choose ε small enough so that max(|S|,|I|)>ε.
we then know that there is α such that |f(x)| < ε for all |x|>α, so we only have to maximise/minimise the function on the compact set [-α,α] which we can do by EVT. Either the max is S or the min is I; this proves the result.
Alternatively, choose $\alpha$ minimally (as an infimum) and assume without loss $f$ is not constant. Suppose for a contradiction that $f$ does not achieve its sup and also does not achieve its inf. If $M\geq\varepsilon$ then M is a global maximum and if $m\leq-\varepsilon$ then m is a global minimum. So the only way for this to happen is if $m$ and $M$ are in $(-ε,ε)$; this leads to a contradiction as ε→0. (see chat/comments for details)
Best Answer
On the request of OP I am converting my comment into an answer.
The result in question is a non-trivial property of continuous functions and it is a consequence of completeness of real numbers. The completeness of real numbers is the only property of real numbers that distinguishes it from the rational numbers and it can be expressed in many equivalent forms and perhaps the most popular one is as follows:
Numbers of the form $K$ are said to be an upper bound of set $A$ and it should be obvious that if $K$ is an upper bound of $A$ then any number greater than $K$ is also an upper bound of $A$. The number $M$ in the above result is called the least upper bound or supremum of $A$ and we write $M=\sup A$.
The terms bounded below, lower bound, greatest lower bound or infimum can be defined similarly and it can be proved easily (using the above mentioned result) that if a non-empty subset of the set of real numbers is bounded below then it has a greatest lower bound or infimum.
Let's assume the following property of continuous functions (which is a consequence of the completeness of real numbers as will be shown later):
Next we come to the Extreme Value Theorem which is the result in question:
This theorem is an easy consequence of theorem 1 mentioned above. Since $f$ is continuous on $[a, b] $ it follows by theorem 1 that it is bounded on $[a, b] $. Let $M, m$ be the supremum and infimum (respectively) of the values of $f$ on $[a, b] $ (these numbers $m, M$ exist via completeness property of real numbers). We will show that there is a point $d\in[a, b] $ such that $f(d) =M$.
Let's assume on the contrary that $f(x) \neq M$ for all $x\in[a, b] $. Then the function $g$ defined by $g(x) =1/(M-f(x))$ is continuous on $[a, b] $ and by theorem $1$ it is bounded on $[a, b] $. But since $M$ is the supremum of values of $f$ there are values of $f$ which can be as near to $M$ as we please so that $g(x) =1/(M-f(x))$ is unbounded. This contradiction proves that $M=f(d) $ for some $d\in[a, b] $. Similarly we can show that $m=f(c) $ for some $c\in[a, b] $.
Theorem 1 stated earlier can be proved easily using completeness of real numbers. Since $f$ is continuous at $a$ there is a $c\in(a, b) $ such that $|f(x) - f(a) |<1$ for all $x\in[a, c] $. Thus $f$ is bounded on $[a, c] $. Consider the following set: $$A=\{x\mid x\in[a, b], f\text{ is bounded on }[a,x] \} $$ It is clear from the above argument that $[a, c] \subseteq A\subseteq [a, b] $ so that $A$ is non-empty and bounded above. By completeness property of real numbers $M=\sup A$ exists and clearly $M\in(a, b] $. We will show that $M\in A$ and $M=b$.
Since $f$ is continuous on $M$, it follows that there is a number $h>0$ such that $[M-h, M]\subseteq [a, b] $ and $f$ is bounded on $[M-h, M]$. Since $M=\sup A$, there is a member $x\in A$ such that $x\in(M-h, M] $ and therefore $f$ is bounded on $[a, x] $. It is now clear that $f$ is bounded on $[a, M] $ so that $M\in A$.
Next let us assume on the contrary that $M<b$. Then as before there is a number $h>0$ such that $[M-h, M+h] \subseteq [a, b] $ and $f$ is bounded on $[M-h, M+h] $. And as before there is an $x\in A$ such that $x\in(M-h, M] $ so that $f$ is bounded on $[a, x] $. It follows that $f$ is bounded on $[a, M+h] $ and thus $M+h\in A$. This is a contradiction as $M=\sup A$. Thus $M=b$ and therefore $f$ is bounded on $[a, b] $.
Both theorem 1 and Extreme Value Theorem can be proved independently using various formulations of completeness property and one should try to prove these results using all the different forms of completenes. This helps in understanding the completeness property as well the properties of continuous functions on closed intervals.