Compactness and closedness 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.
 A: 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 ${\mathcal C}$.
A: My suggestion:
Let $T$ be the topological space. Let $U_{i}\neq\emptyset, i\in I$ be an open cover.
Let $i_{0}\in I$ and consider $T^{'}:=T\setminus U_{i_{0}}$. $T^{'}$ is a proper subspace and thus there is a finite subset $I_{F}\subset I$ such that $\left\{U_{i}\right\}_{i\in I_{F}}$ is an open cover for $T^{'}$ but this means that $\left\{U_{i}\right\}_{i\in I_{F}\cup\left\{i_{0}\right\}}$ is a finite open cover for $T$.
Editet it twice, in this version I don't need Zorn's lemma. Check if I forgot something plx :-)
