It's known to be consistent with ZF+DC that every subset of $\mathbb{R}$ has the Baire property (BP).   (E.g. Shelah's model).  If so, then every subset of every complete separable metric space has the BP.

Can we drop the word "separable" here? 

> Is it consistent with ZF+DC that every subset of every complete metric space has the BP?

In other words, working in ZF+DC, can we prove there exists a complete (non-separable) metric space $X$ and a subset $E \subset X$ without the BP?  

I'm not sure which way my intuition goes.  On the one hand, non-separable metric spaces are big and maybe they can be weird, even without AC.  On the other hand, a counterexample would have the property that the intersection of $E$ with every separable $S \subset X$ would (consistently) have the BP in $S$.  That seems unlikely but I don't see how to disprove it.

As one possible approach, a nonmeager proper linear subspace of any Banach space $B$ must not have the BP (by the Pettis lemma).  Can we create such a beast in ZF+DC, for some non-separable $B$?  We can't do it by taking the kernel of a discontinuous linear functional $f$, because it's consistent with ZF+DC that there aren't any (by considering sequences, the restriction of $f$ to some closed separable subspace $S$ would be discontinuous, making its kernel nonmeager in $S$).