Let \(A\subset X\) with \(A\) closed and \(X\) compact, and show that \(A\) is compact.
Solution
Let \(\mathcal U\) be an open cover of \(A\) by subsets open in the subspace topology. For each \(U\in\mathcal U\), choose an open set \(V_U\subseteq X\) with \(U=A\cap V_U\). Since \(A\) is closed, \(X\setminus A\) is open, and \[ \{V_U:U\in\mathcal U\}\cup\{X\setminus A\} \] is an open cover of \(X\). Compactness of \(X\) gives a finite subcover \[ V_{U_1},\ldots,V_{U_n},X\setminus A. \] Intersecting with \(A\) removes the last set and gives \[ A=(A\cap V_{U_1})\cup\cdots\cup(A\cap V_{U_n}) =U_1\cup\cdots\cup U_n. \] Thus every open cover of \(A\) has a finite subcover, so \(A\) is compact.