Let $B$ be a topological space.
Call a subset $A\subset B$ *ultrafilter-like* iff $A$ is dense in $B$ and
each decomposition $A=A_1\cup A_2$,
into the union of two open subsets extends to a decomposition of
$B=B_1\cup B_2$, $A_1\subset B_1$ and $A_2\subset B_2$ into the union of two open subsets $B_1$ and $B_2$.

Is the following true? Let $f:X\rightarrow Y$ be a proper map and $A\subset B$ is ultrafilter-like. For any $g:A\rightarrow X$ and $h:B\rightarrow Y$ such that $f(g(a))=h(a)$ for each $a\in A$, there is a map $g':B\rightarrow X$ extending $g$ such that $f(g'(b))=h(b)$ for each $b\in B$.

Is this true if $Y$ is a point (and thus $X$ is an arbitrary quasi-compact space).

This is true provided $B=A\cup \{\omega\}$ where $\omega$ is closed. If $A$ is discrete, then the neighbourhoods of $\omega$ define an ultrafilter on $A$ and the property becomes the definition of a proper map via ultrafilters (see Bourbaki, General Topology, I\S10.2, Theorem I). If $A$ is not necessarily discrete, then there is an ultrafilter $F$ on $A$ such that the neighbourhoods of $\omega$ are $F$-big, and the same arguments works. (To fix the terminology: by a proper map I mean a closed map such that the preimage of any point is quasi-compact).