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.