Let $X$ be a locally compact Hausdorff space and $Y = X \cup \{\infty\}$ its one point compactification. The following is Munkres' proof that $Y$ is compact.
Let $\mathscr{A}$ be an open covering of $Y$. The collection $\mathscr{A}$ must contain an open set of type $Y- C$ (where $C$ is compact in $X$), since none of the open sets contained in $X$ contain the point $\infty$. Take all the members of $\mathscr{A}$ different from $Y-C$ and intersect them with $X$; they form a collection of open sets in $X$ covering $C$. Because $C$ is compact, finitely many of them cover $C$; the corresponding finite collection of elements of $\mathscr{A}$ will, along with the element $Y - C$, cover all of $Y$.
I have two questions about his proof.
- For each $U \in \mathscr{A} \setminus (Y-C)$ how do we know that each intersection $U \cap X$ belongs to $\mathscr{A}$, as is needed to obtain a subcover?
- Why does it suffice to only consider one set of the type $Y - C$? What if $\mathscr{A}$ has uncountably many sets containing the point at infinity?
Regarding question 1: The set $U \cap X$ does not necessarily belong to $\mathscr{A}$.
But you do not need it to belong to $\mathscr{A}$.
All you need is is to find finitely many $U$'s in $\mathscr{A}$, denote them $U_1,...,U_m$, such that $C \subset U_1 \cup \cdots \cup U_m$, in other words such that $U_1,...,U_m$ cover $C$.
And the way you find those $U$'s is to apply compactness of $C$, like this. The set $\{U \cap C \mid U \in \mathscr A\}$ is an open cover of $C$ with respect to the subspace topology on $C$. By compactness of $C$ finitely many of those open sets form a cover of $C$: $$C = (U_1 \cap C) \cup \cdots \cup (U_m \cap C) $$ By elementary set theory, it follows that $C \subset U_1 \cup\cdots\cup U_m$.
Regarding question 2: Here's why it suffices to only consider one set of the type $Y-C$. You have already picked out $Y-C \in \mathscr A$ itself. And after that, you have already picked out finitely many elements of $\mathscr A$ that cover $C$, namely $U_1,...,U_m$. So now you just need finitely many more elements of $\mathscr A$ that cover everything in $Y$ BUT $C$. That is, you just need finitely many more elements of $\mathscr A$ that cover $Y-C$.
Hurray!! $Y-C$ is a single element of $\mathscr A$ that covers $Y-C$!!