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?
 A: 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).
(This construction more generally shows that any infinite Boolean algebra has an infinite sequence of nonzero pairwise disjoint elements.)
