Suppose A and B are Hausdorff compacta and $f:A \to B $ is a function.
Note if $f$ is a continuous function then
$1.$ $f^{-1}(b)$ is closed in A for each b in B.
$2.$ $f(C)$ is compact for each compact subspace $C$ of $A$.
Do conditions $1$ and $2$ characterize continuity?
Yes.
Suppose that $f:A\to B$ has properties $(1)$ and $(2)$; we may as well also assume that $f$ is surjective. Let $a\in A$, and let $\mathscr{H}$ be the set of closed nbhds of $a$; $A$ is regular, so $\mathscr{H}$ is a filterbase for the nbhd filter at $a$. Let $\mathscr{G}=\{f[H]:H\in\mathscr{H}\}$; $\mathscr{G}$ is a filterbase of closed sets in $B$, and $f(a)\in\bigcap\mathscr{G}$.
For each $x\in A$ let $F_x=f^{-1}[\{f(x)\}]$. If $x\in A\setminus F_a$, then $F_a$ and $F_x$ are disjoint closed sets in $A$, so there is an $H\in\mathscr{H}$ such that $H\cap F_x=\varnothing$, and hence $f(x)\notin f[H]$. Thus, $\bigcap\mathscr{G}=\{f(a)\}$.
If $f$ is not continuous at $a$, there is an open nbhd $U$ of $f(a)$ such that $f[H]\nsubseteq U$ for each $H\in\mathscr{H}$. Let $K=B\setminus U$; clearly $K\cap\bigcap\mathscr{G}=K\cap\{f(a)\}=\varnothing$. On the other hand, $f[H]\cap K\ne\varnothing$ for each $H\in\mathscr{H}$, so $\{K\}\cup\mathscr{G}$ is a centred family of closed sets (i.e., it has the finite intersection property), and therefore $K\cap\bigcap\mathscr{G}\ne\varnothing$, since $B$ is compact. This contradiction shows that $f$ must be continuous at $a$, and since $a\in A$ was arbitrary, $f$ is continuous.
