closed set in a compact space is compact

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 hasa 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 thatcovers $X$. Let $D$ be this subcover.Either $A^c$ is part of $D$ or $A^c$ is not.In any case, $D\backslash \backslash \{A^c\}$ is a finite open coverfor $A$, and $D\backslash \backslash \{A^c\}$is a subcover of ${\{U_i\}}_{i\in I}$. The claim follows. $\mathrm{\square }$