Question: Let $X$ and $Y$ be topological spaces, $X$ compact, $f : X \to Y$ continuous. Then the preimage of each compact subset of $Y$ is compact.

With the stipulation that $X$ and $Y$ are metric spaces, this is a theorem in Pugh's *Real Mathematical Analysis*. The proof uses sequential compactness. Is this theorem true in general (*i.e.* can it be proved with covering compactness alone)?

Answer: 
The claim is true if $Y$ is Hausdorff: if $C\subseteq Y$ is compact, then it is closed; therefore $f^{-1}(C)$ is closed in $X,$ hence compact.

For a counterexample, take $Y=X$ with the indiscrete topology and $f$ the identity map. Then every subset of $Y$ is compact, which can be easily arranged for $X$ not to.
**Disclaimer:** This answer is intended to compliment the above, and thought it is not fully generally, but is has often been helpful in my experience... 

If $Y$ is $\mathbb{R}$ then its sufficient for $f$ to be [coercive](https://en.wikipedia.org/wiki/Coercive_function). When $X$ is a topological vector space, the coercivity condition reduces to the requirement that
$$
\lim\_{\|x\|\to \infty} f(x)=\infty,
$$
which is typically easy to check.