Let ${\displaystyle f:X\to Y}$ be a closed map, such that ${\displaystyle f^{-1}(y)}$ is compact (in ${\displaystyle X}$) for all ${\displaystyle y\in Y.}$ Let ${\displaystyle K}$ be a compact subset of ${\displaystyle Y.}$ It remains to show that ${\displaystyle f^{-1}(K)}$ is compact.

Let ${\displaystyle \left\{U_{a}:a\in A\right\}}$ be an open cover of ${\displaystyle f^{-1}(K).}$ Then for all ${\displaystyle k\in K}$ this is also an open cover of ${\displaystyle f^{-1}(k).}$ Since the latter is assumed to be compact, it has a finite subcover. In other words, for every ${\displaystyle k\in K,}$ there exists a finite subset ${\displaystyle \gamma _{k}\subseteq A}$ such that ${\displaystyle f^{-1}(k)\subseteq \cup _{a\in \gamma _{k}}U_{a}.}$ The set ${\displaystyle X\setminus \cup _{a\in \gamma _{k}}U_{a}}$ is closed in ${\displaystyle X}$ and its image under ${\displaystyle f}$ is closed in ${\displaystyle Y}$ because ${\displaystyle f}$ is a closed map. Hence the set $${\displaystyle V_{k}=Y\setminus f\left(X\setminus \cup _{a\in \gamma _{k}}U_{a}\right)}$$ is open in ${\displaystyle Y.}$ It follows that ${\displaystyle V_{k}}$ contains the point ${\displaystyle k.}$ Now ${\displaystyle K\subseteq \cup _{k\in K}V_{k}}$ and because ${\displaystyle K}$ is assumed to be compact, there are finitely many points ${\displaystyle k_{1},\dots ,k_{s}}$ such that ${\displaystyle K\subseteq \cup _{i=1}^{s}V_{k_{i}}.}$ Furthermore, the set ${\displaystyle \Gamma =\cup _{i=1}^{s}\gamma _{k_{i}}}$ is a finite union of finite sets, which makes ${\displaystyle \Gamma }$ a finite set.

Now it follows that ${\displaystyle f^{-1}(K)\subseteq f^{-1}\left(\cup _{i=1}^{s}V_{k_{i}}\right)\subseteq \cup _{a\in \Gamma }U_{a}}$ and we have found a finite subcover of ${\displaystyle f^{-1}(K),}$ which completes the proof.