Here's a trivial answer on the second question. Yes.
For two counts:
It is consistent with $\sf ZF$ that $\sf AC$ is true. Therefore $\frak BS$ is a singleton, so it is closed under "intersections".
Okay, so the above was a bit of a cheat, because we are clearly care about the case where the axiom of choice fails. Still, Sizes of bases of vector spaces without the axiom of choice shows that assuming $\sf BPI$ we have that every two bases have the same cardinality. Therefore it is consistent that the only values that $\frak BS$ takes (and it takes them both) are $\varnothing$ or singletons.