False: $\left\{1-\frac1n\right\}_{n\in\Bbb N}$.
Does it has something to do with closure? Yes, a closed bounded set necessarily contains its supremum, even if it is uncountable.
Suppose $A \subset \Bbb{R}$, non-empty, bounded above, and closed.
Then, $\sup(A)$ exists.
Suppose in order to derive a contradiction that $\sup(A) \notin A$.
Then, $\sup(A) \in \Bbb{R} \setminus A$.
Since $A$ is closed, $\Bbb{R} \setminus A$ is open.
Thus, $\Bbb{R} \setminus A$ entirely contains $B(\sup(A), \varepsilon)$, the ball of center $\sup(A)$ and of some radius $\varepsilon > 0$.
Then, $B(\sup(A), \varepsilon)$ contains the number $\sup(A) -\frac{\varepsilon}{2}$.
Since $\sup(A) - \frac{\varepsilon}{2} \in B(\sup(A), \varepsilon)$, $\sup(A) - \frac{\varepsilon}{2} \geq x, \forall x \in A$.
But $\sup(A) - \frac{\varepsilon}{2} < \sup(A)$.
This contradicts our assumption that $\sup(A)$ was the least upper bound.
So, we must have $\sup(A) \in A$.
Best Answer
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.