Solution: Let $(X,\mathcal{T} )$ be a compact topological space and $Y \subseteq X$ be closed. We need to prove that $Y$ is compact. Let $\mathcal{U} = \{ U_\alpha : \alpha \in \mathcal{I} \} $ be an open cover of $Y$, where $\mathcal{I} $ is an index set. Since $Y$ is closed, $X\setminus Y$ must be open. Also, for each $\alpha \in \mathcal{I} $ the set $U_\alpha $ is open in $X$, so \[ \mathcal{U} \cup (X\setminus Y) = \{ U_\alpha : \alpha \in \mathcal{I} \} \cup (X\setminus Y) \] is an open cover of $X$. As $X$ is compact, the above cover admits a finite subcover. Let \[ X \subseteq U_1 \cup U_2 \cup \dots U_k \cup (X\setminus Y). \] Note that if $X\setminus Y$ is not in the finite subcover, then we can include it and still it is a finite subcover. Therefore, \begin{align*} Y \subseteq X \subseteq U_1 \cup U_2 \cup \dots U_k \cup (X\setminus Y), \end{align*} which implies \[ Y \subseteq U_1 \cup U_2 \cup \dots U_k \] as $Y \cap (X\setminus Y) = \emptyset $. Therefore, $Y$ is compact.