Question

I think you can have such a subspace. Let $f : X \to R R$ be a discontinous linear functional (such a functional exists assuming Axiom of Choice, see wikipedia). The claim is that its kernel $K = ker(f) \ker f$ is a proper non-meager subspace. It is definitely proper. Assume it would be meager. Then it is contatined contained in the countable union of closed subsets Ai. $A_i$. Since K=ker(f) $K=\ker f$ it has codimesnion codimension 1, so there is $z \in X X$ such that every $x \in X X$ can be written as $x = k + azaz$, for some number a $a$ and some $k \in KK$. Let Bi $B_i$ be the set of elemets Ai elements $A_i + [-i,i]z -i,i]z$ (that is the set of $x \in X X$ for whcih which $x = k+az we have k k+az$ where $k$ is in Ai $A_i$ and a $a$ in [-i,i]). $[-i,i]$). Then, I think, Bi $B_i$ are closed and they have to have empty enteriorinterior. Indeed, if there is a small ball around $k + az az$ in Bi $B_i$ then f $f$ will be continuous at k+az, $k+az$, and then (since f $f$ is linear) continuous everywhere, contradicting the choice of f. $f$. Thus Bi $B_i$ are closed and nowhere dense, but their union is then the whole space X, $X$, which contradicts Baire's theorem.