"Closed subsets of compact space are compact" with an alternative definition of covering I am studying Abraham and Marsden's Foundations of Mechanics. They define a covering $\{U_\alpha\}$ of a set $S$ to satisfy $S = \bigcup_\alpha U_\alpha$, and NOT as $S\subseteq\bigcup_\alpha U_\alpha$.
Their definitions of compactness of a topological space, and its subsets (involving the relative topology) are the usual ones.
Then they casually remark

It follows easily that a closed subset of a compact space is compact.

However, I'm not able to prove this with the given definition of cover. (I can prove it using the latter definition though.)
I'm even skeptical if that holds with the given definition.

I really don't think that this is a duplicate!
 A: 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 $S$.
