Proof. Let $A$ be a closed set in a compact space $X$. To show that $A$ is compact, we show that an arbitrary open cover has a finite subcover. For this purpose, suppose $\{U_{i}\}_{{i\in I}}$ be an arbitrary open cover for $A$. Since $A$ is closed, the complement of $A$, which we denote by $A^{c}$, is open. Hence $A^{c}$ and $\{U_{i}\}_{{i\in I}}$ together form an open cover for $X$. Since $X$ is compact, this cover has a finite subcover that covers $X$. Let $D$ be this subcover. Either $A^{c}$ is part of $D$ or $A^{c}$ is not. In any case, $D\backslash\{A^{c}\}$ is a finite open cover for $A$, and $D\backslash\{A^{c}\}$ is a subcover of $\{U_{i}\}_{{i\in I}}$. The claim follows. $\Box$