Let $X$ be a Tychonoff space with no isolated points such that the boundary of any subset of $X$ is compact. Does it mean that $X$ is compact ? (If $X$ is a resolvable space then it is clearly compact.) 



When the boundary of any subset is compact?
Your property implies in particular:


"Every nowhere dense closed subset of $X$ is compact."


(A nowhere dense closed subset is the boundary of its complement.)


This in turn is equivalent to compactness for $T\_1$ spaces with no isolated points, as shown by Katetov in 1947. Interestingly, the result holds as well for Lindelöf spaces, or more generally for $[\lambda,\kappa]$-compact spaces (a space is $[\lambda,\kappa]$-compact if every cover of it by $\le\kappa$ open sets has a subcover of cardinality $<\lambda$). This is due to Mills and Wattel, a very nice short proof can be found in Blair's "Some nowhere densely generated topological properties" (easily found online).


Completely unaware of these results, I actually recently published a paper whose main result is weaker than Mills and Wattel's, and with a more complicated proof than Blair's.


