Solution: Let $F$ be a closed subset of $X$. We need to show that $F$ is compact. Take an open cover of $F$, say $\mathcal{U} = \left\{ U_\alpha \right\}_{\alpha \in I}$, where $I$ is an index set. We will show that $\mathcal{U} $ admits a finite subcover. Since $F$ is closed, the complement $X\setminus F$ must be open in $X$ and hence $\mathcal{U} \cup \left( X\setminus F \right) $ will be an open cover for $X$. Since $X$ is compact, this cover admits a finite subcover. Assume that \[ X = \bigcup_{i=1}^{n} U_{\alpha _i} \cup \left( X\setminus F \right). \] Therefore, \[ F \subseteq \bigcup_{i=1}^{n} U_{\alpha _i} \] and hence, $\mathcal{U} $ admits a finite subcover which proved that $F$ is compact.