Proof
(related to Proposition: Closed Subsets of Compact Sets are Compact)
 Let $(X,d)$ be a metric space and let \(C\subset X\) be a compact subset of \(X\).
 It has to be shown that any closed subset \(A\subset C\) is also compact.
 Let $U_i)_{i\in I}$ be an open cover of $A$.
 Since \(A\) is closed, the set \(X\setminus A\) is open.
 Thus, $(X\setminus A)\cup\bigcup_{i\in I}U_i=X$ is an open cover of $C$.
 Since \(C\) is compact, it has by definition of compact sets a finite subcover $$C\subset (X\setminus A)\cup U_{i_1}\cup U_{i_2}\cup\ldots\cup U_{i_k}$$ for some \(i_1,i_2,\ldots,i_k\in I\).
 Because \(A\subset C\), also \(A\) has a finite subcover.
 Thus, \(A\) is compact.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

References
Bibliography
 Forster Otto: "Analysis 2, Differentialrechnung im \(\mathbb R^n\), Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984