TITLE: Compactness and closedness
QUESTION [6 upvotes]: If every closed and proper subset of a topological space $X$ is compact, then is the whole space necessarily compact? 
The "converse" of this question is well-known, of course, but I'm having difficulty establishing a proof of this. Also, no counterexamples spring to mind either.

REPLY [6 votes]: An equivalent definition of compactness is the following:
A space $X$ is compact if and only if every family of closed subsets of $X$ with the finite intersection property has non-empty intersection.
We say that a family $\mathcal F$ of sets has the finite intersection property if 
$F_1\cap\cdots\cap F_n\ne\varnothing$,
for every $n$ and $F_1,\ldots,F_n\in\mathcal F$. (See also here.)
Assume now that $\mathcal C$ is a family of closed subsets of $X$ with the finite intersection property, and $F\in\mathcal C$, with $F\ne X$. It is already given that $F$ is compact. Then clearly the family
$$
\tilde{\mathcal C}\,=\,\big\{F\cap C: C\in\mathcal C\big\},
$$
is another family of closed subsets of $X$ with the finite intersection property, and as they are also closed subsets of $F$, which is assumed compact, 
the family $\tilde{\mathcal C}$ has non-empty intersection, and so does 
family ${\mathcal C}$.