If a topological space can be partitioned into finitely many second countable subspaces, is it second countable? [Collecting examples]

Let X be the result of glueing countably infinitely many copies of [0,1] together at 0. It is not second countable, it even fails to be first countable at 0. But {0} and X - {0} are each second countable.

(ETA) Suppose Y is an uncountable, second countable, T1 space and Z a countable, discrete space, with Y and Z disjoint. Let X be their union with the topology with a basis comprising the open sets of Y and sets of the form C∪{z} with z∈Z and C a cofinite subset of Y. X is not first countable as no point of Z has a countable neighbourhood basis in X, but it splits into Y and Z which are both second countable.

Both examples depend on X being not first countable, which suggests the follow-up question: If a first countable space can be partitioned into two second countable subspaces, is it second countable?