If \(X\) is compact, and \(A\subseteq Y\) is closed, then \(A\) is compact.
Solution
The statement assumes \(X\) is compact but writes \(A\subseteq Y\); we prove it with \(A\subseteq X\): if \(X\) is compact and \(A\subseteq X\) is closed, then \(A\) is compact.
1. Let \(\mathcal U\) be an open cover of \(A\) by sets open in \(X\).
To prove compactness of \(A\), it is enough to show that every open cover of \(A\) admits a finite subcover. Any open cover by sets open in the subspace \(A\) can first be written in the form \(A\cap U\) with \(U\) open in \(X\), so it suffices to treat covers by ambient open sets.
2. The family \(\mathcal U\cup\{X\setminus A\}\) is an open cover of \(X\).
Because \(A\) is closed in \(X\), the complement \(X\setminus A\) is open. The family \(\mathcal U\) covers \(A\), while \(X\setminus A\) covers every point of \(X\) not in \(A\).
3. There are finitely many sets \(U_1,\dots,U_n\in\mathcal U\) such that\[ X\subseteq U_1\cup\cdots\cup U_n\cup(X\setminus A). \]
2.If a finite subcover does not use \(X\setminus A\), the same conclusion holds after adjoining it.
4. The sets \(U_1,\dots,U_n\) cover \(A\).
Intersect the inclusion in 3 with \(A\). Since \(A\cap(X\setminus A)=\varnothing\), one obtains \[ A\subseteq U_1\cup\cdots\cup U_n. \]
5. Therefore \(A\) is compact.
The open cover \(\mathcal U\) in 1 was arbitrary, and 4 produced a finite subcover.