Closed subsets of a compact topological space are compact proof clarification 3

Question

I have come across the following proof of the stated theorem:

I fail to understand somehow why this depends on $K$ being closed? Can we just argue that $K \subseteq X \subseteq \cup_n U_n$ ($\{U_n\}$ finite open cover of $X$) since $X$ compact? This would easily generalise to any subset of $X$ which seems wrong.

Answer 1

It depends on $K$ being closed because then $X\setminus K$ is open. Thus, when it is included in the cover, it is truly an open cover, of which you may take a finite subcover.

Note the method is this: they take an arbitrary open cover of $K$. By adding a single open set (that is, $X\setminus K$), we extend it to all of $X$. Since $X$ is compact, there is a finite subcover of the cover that we just listed. You can use this to deduce that $K$ is covered by finitely many elements in the subcover (which came from an arbitrary open cover). In particular, $K$ must be compact.

Answer 2

Just to review the definition of compactness, it says that that FOR ANY collection of open subsets $\{U_\gamma\}_{\gamma \in \Gamma}$ which covers $K$ (meaning that $K \subset \bigcup_{\gamma \in \Gamma} U_\gamma$), there exists a finite number of them $U_{\gamma_1},...,U_{\gamma_n}$ which cover $K$ (meaning that $K \subset \cup_{i=1}^n U_{\gamma_n}$).

In your question, you are restricting your attention to a collection of open sets which covers $X$. But that's not enough, you must consider ANY collection of open sets that covers $K$, not just ones that cover $X$.