Solution: Since $A$ is non-empty bounded subset of $\mathbb{R}$, the supremum and infimum of $A$ will exist. Let us denote \[ \sup A = \alpha \quad \text{ and } \quad \inf A = \beta . \] We need to prove that $\alpha ,\beta \in A$.

Since $\alpha $ is the supremum of $A$, for any $\epsilon >0$, we can find an element $x \in A$ such that \[ \alpha -\epsilon \lt x \lt \alpha \implies \alpha -\epsilon \lt x \lt \alpha + \epsilon . \] This implies that any ball $B(\alpha ,\epsilon ) \cap A \neq \emptyset $. So, $\alpha $ is a limit point of $A$. As $A$ is closed, it must contain the limit point and hence, $\alpha \in A$.

Similarly, for any $\epsilon >0$, we can find $y\in A$ such that \[ \beta \lt y \lt \beta + \epsilon \implies \beta - \epsilon \lt y \lt \beta +\epsilon . \] Hence, by the same argument, $\beta $ is a limit point of $A$ and thus, $\beta \in A$.