"Closed subsets of compact space are compact" with an alternative definition of covering

Solution 1:

The key idea is to "lift" each element of your cover of $S$ to a cover of your ambient space. Since you seem to be struggle to intuit how that works from the usual proof, here is a full proof with details.

Let $\{U_{\alpha}\}_{\alpha}$ be an open cover of $S\subseteq X$; that is, each $U_{\alpha}$ is open in the subspace topology on $S$ induced by $X$, and such that $$\bigcup_{\alpha}{U_{\alpha}}=S$$

Fix $\alpha$. Since $U_{\alpha}$ is open in the subspace topology induced by $X$, there exists an set $V_{\alpha}\supseteq U_{\alpha}$ such that $V_{\alpha}\subseteq X$ is open (in the natural topology on $X$) and $V_{\alpha}\cap S=U_{\alpha}$.

Now vary $\alpha$, and consider the collection made by combining $\{V_{\alpha}\}_{\alpha}$ with $\{X\setminus S\}$. Each element in this collection is open in $X$ and \begin{align*} X&=(X\setminus S)\cup S \\ &=(X\setminus S)\cup\bigcup_{\alpha}{U_{\alpha}} \\ &\subseteq(X\setminus S)\cup\bigcup_{\alpha}{V_{\alpha}} \\ &\subseteq X\cup\bigcup_{\alpha}{X} \\ &=X \end{align*} Thus all these sets are equal. In particular, $$X=(X\setminus S)\cup\bigcup_{\alpha}{V_{\alpha}}$$

So our new collection is an open cover of $X$. Since $X$ is compact, it has a finite subcover; say, $\{V_{\beta}\}_{\beta}$, possibly combined with $\{X\setminus S\}$. But then \begin{align*} S&=X\cap S \\ &=((X\setminus S)\cup\bigcup_{\beta}{V_{\beta}})\cap S \\ &=((X\setminus S)\cap S)\cup\bigcup_{\beta}{(V_{\beta}\cap S)} \\ &=\emptyset\cup\bigcup_{\beta}{U_{\beta}} \\ &=\bigcup_{\beta}{U_{\beta}} \end{align*} Thus $\{U_{\beta}\}_{\beta}$ is a finite open cover of $S$.