Here's a proof that any closed subset of a compact metric space is compact -
Let $(X, d)$ be a compact metric space and let $F$ be a closed subset of $X$.
Let $U = \{U_i : i \in I \}$ be an open cover of $F$.
Now we can form $V = \{X\backslash F\} \cup U$ which is an open cover of $X$.
As $X$ is compact $\exists$ finitely many $V_is$ from $V$ such that $\cup_i^n V_i = X$
Those $V_is$ from $U$ will cover $F$ and hence F is compact.
Question
What is the significance of F$ being closed, I don't see any mention of that fact being made in the proof? If we had $F$ is an open subset where would the proff