Does a closed and bounded set in $\mathbb{R}$ necessarily contain its supremum and infimum?

The proof is correct, but it's rather non-minimal, since the extreme value theorem you use is stronger than the result you want and the proof for this specific case is easier than the proof of that theorem.

Assume the set didn't contain its supremum (which exists since the set is bounded). Since the set is closed, an entire neighbourhood of the supremum lies outside the set. This neighbourhood contains upper bounds of the set that are lower than the supremum, which is a contradiction.

Here is another way of proving the result, which is (was) similar to joriki's answer.

Let $A \subset \mathbb{R}$ be closed and bounded. If $A$ is a singleton, then we are done. As long as $A$ has a finite cardinality this argument holds. If $A$ has infinitely many elements, let $L=\sup A$, which exists since $A$ is compact and $\mathbb{R}$ is complete. The idea now is to construct a sequence in $A$ converging to $L$, from whence the claim would then follow. Take an $\epsilon > 0$, then there exists an $x_\epsilon \in A$ such that

$$\tag{1} L-\epsilon \le \ x_\epsilon \le L $$

by definition of the supremum. Since $\epsilon$ is arbitrary, we can take the limit as it approaches 0.

We would need to consider

$$ \underset{\epsilon \to 0}{\lim}\ L-\epsilon \le \underset{\epsilon \to 0}{\lim}\ x_\epsilon \le L $$ We argue that there exists an $x_\epsilon$ which still belongs to $A$ for each $\epsilon$ chosen by asserting equation (1).

This gives us $$L\le x_0 \leq L \quad \text{where} \quad x_0= \underset{\epsilon \to 0}{\lim}\ x_\epsilon $$ Thus $$L \in A$$ Namely, $A$ contains $L=x_0$ since it is a limit point, and $A$ is closed.