$q:X\to Y$ continuous, show that $X$ is compact I've been stuck on this problem for a very long time:
Problem: Let $X,Y$ be topological spaces, $Y$ compact, and let $q:X\to Y$ be a continuous and closed map. Assume that $q^{-1}(\{y\})$ is compact for every $y\in Y$. Show that $X$ is compact.
What I have so far:
$q:X\to q(X)$ is also continuous and closed, and $q(X)$ is compact (as a closed subset of a compact space). Thus, we could assume that $q(X)=Y$, or that $q$ is onto $Y$.
If $q^{-1}(A)\subset X$ is open, Then $(q^{-1}(A))^c=q^{-1}(A^c)$ is closed, and so ($q$ is onto) $A^c=q(q^{-1}(A^c))$ is closed and $A$ is open. Thus $q$ is a quotient map.
Thus, the quotient space $X^*=\{q^{-1}(\{y\})|y\in Y\}$ is compact, as it's homeomorphic to $Y$.
How do I carry on from here? 
 A: I found this proof in some notes of mine.  I'm sure it's not original but I don't know the source.
Let $\mathcal A$  be a family of closed subsets of $X$ with the finite intersection property.  We need to show that $\bigcap_{A \in \mathcal A} A \ne \emptyset$. 
We can suppose wlog that this family is closed under finite intersections, because augmenting $\mathcal A$ with all finite intersections of its elements preserves both the FIP and  $\bigcap_{A \in \mathcal A} A $.
Consider 
$\{q(A)\}_{A \in \mathcal A}$;  this is a family of closed subsets of $Y$ since $q$ is a closed map.  Show that this family has the finite intersection property.   Since $Y$ is compact there exists a point $y \in \bigcap_{A \in \mathcal A} q(A)$. 
Equivalently, $q^{-1}(y) \cap A  \ne \emptyset$ for all $A \in \mathcal A$.
The family $\{q^{-1}(y) \cap A\}_{A \in \mathcal A}$ has the FIP because for $A_i \in \mathcal A$,
$$
(q^{-1}(y) \cap A_1) \cap \cdots \cap (q^{-1}(y) \cap A_k)  = q^{-1}(y) \cap (A_1 \cap \cdots \cap A_k),
$$
$A_1 \cap \cdots \cap A_k \in \mathcal A$ by the assumption that $\mathcal A$ is closed under finite intersections, and we have just shown that $q^{-1}(y) \cap A  \ne \emptyset$ for all $A \in \mathcal A$.
But the family $\{q^{-1}(y) \cap A\}_{A \in \mathcal A}$ is a family of closed subsets of the compact set 
 $q^{-1}(y)$.  Hence 
 $\bigcap_{A \in \mathcal A} (q^{-1}(y) \cap A) \ne \emptyset$.
Notes:  Continuity of $q$ is not used in the proof.   However,  the result is a lemma for the following proposition.    Call a map $f: X \to Y$  between topological spaces \emph{proper} if the inverse image 
 of every compact subset of $Y$ is compact in $X$.   
Proposition:  Let $X$ and $Y$ be topological spaces with $Y$ Hausdorff.  Let 
 $f: X \to Y$  be continuous and closed with each fiber $f^{-1}(y)$ compact in $X$.  Then $f$ is proper.
Proof:  Let $K$ be compact in $Y$.  Then $K$ is closed since $Y$ is Hausdorff, and $f^{-1}(K)$ is closed in $X$ since $f$ is continuous.  Let $q$ denote the restriction of $f$ to $f^{-1}(K)$.  The hypotheses of the original question apply to $q$, and hence $f^{-1}(K)$ is compact.
