# For what spaces does every countable Borel cover have a finite subcover?

A topological space $$X$$ is countably compact if every countable open cover of $$X$$ has a finite subcover. But I'm interested in a stronger condition. Suppose that every countable cover of $$X$$ by Borel sets has a finite subcover. Then what properties must X have?

And what are examples of spaces that do and don't have this property?

This is an extraordinarily strong property: it is equivalent to $$X$$ having only finitely many open sets. Indeed, it is easy to see that if $$X$$ has finitely many open sets, then it has only finitely many Borel sets, and so your condition holds trivially.
Conversely, suppose $$X$$ has infinitely many open sets. Then in particular $$X$$ has infinitely many Borel sets. Let us call a set $$A\subseteq X$$ large if $$A$$ is Borel and has infinitely many Borel subsets. Note that if $$A$$ is large and $$A$$ is the disjoint union of two Borel sets $$B$$ and $$C$$, then at least one of $$B$$ and $$C$$ is large, since every Borel subset of $$A$$ is the union of a Borel subset of $$B$$ and a Borel subset of $$C$$. It follows that any large set has a large proper subset (pick some nonempty proper Borel subset, and either it or its complement must be large).
We can now use this to construct a countable Borel cover of $$X$$ with no finite subcover. Starting with $$A_0=X$$, we can pick a large proper subset $$A_1\subset A_0$$, and then a large proper subset $$A_2\subset A_1$$, and so on. The sets $$A_0\setminus A_1, A_1\setminus A_2,\dots$$ and $$\bigcap_n A_n$$ are then all Borel and cover $$X$$, and have no finite subcover (they are disjoint and all except possibly $$\bigcap_n A_n$$ are nonempty).