Proof of Closed Subset of a Compact Space is Compact

Let $A\subseteq X$ be closed, and assume that $X$ is compact. By Compact Subset Criterion via Open Covers in the Ambient Space, it is enough to prove that every open cover of $A$ in $X$ has a finite subcover.

Let $I$ be a set, and let $(U_i)_{i\in I}$ be an open cover of $A$ in $X$. By Open Cover and Subcover of a Subset of a Topological Space, each $U_i$ is an open subset of $X$ and

Since $A$ is closed in $X$, the complement $X\setminus A$ is open in $X$.

Choose a symbol $\ast$ not belonging to $I$, and set

Define a family $(V_j)_{j\in J}$ of subsets of $X$ by

Then every $V_j$ is open in $X$. We claim that $(V_j)_{j\in J}$ is an open cover of $X$ in $X$. Let $x\in X$. If $x\in A$, then since $(U_i)_{i\in I}$ covers $A$, there exists $i\in I$ such that $x\in U_i=V_i$. If $x\notin A$, then by the definition of complement one has $x\in X\setminus A=V_{\ast}$. Hence

So $(V_j)_{j\in J}$ is an open cover of $X$ in $X$.

Apply Compact Subset Criterion via Open Covers in the Ambient Space with the subset $X\subseteq X$. Since $X$ is compact, there exist a natural number $n\in\mathbb{N}$ and elements $j_1,\dots,j_n\in J$ such that

Discard every index equal to $\ast$. The remaining sets are some $U_{i_1},\dots,U_{i_m}$ with $m\in\mathbb{N}$, and we still have

since points of $A$ do not lie in $X\setminus A=V_{\ast}$. Therefore $(U_{i_1},\dots,U_{i_m})$ is a finite subcover of $A$.

We have shown that every open cover of $A$ in $X$ has a finite subcover. By Compact Subset Criterion via Open Covers in the Ambient Space, the subset $A$ is compact in $X$.