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.