Are proper linear subspaces of Banach spaces always meager? Let X be a Banach space, and let Y be a proper non-meager linear subspace of X. If Y is not dense in X, then it is easy to see that the closure of Y has empty interior, contradicting Y being non-meager. So Y must be dense. If Y has the Baire property, then it follows from Pettis Lemma that Y is open and hence closed (since the complement of Y is the union of translates of Y), contradicting Y being proper. Thus, Y must be dense and not have the Baire property.  
My question is: is there a Banach space X with a proper non-meager linear subspace Y? Such a Y must be dense and not have the Baire property. Any such Y must be difficult to construct since all Borel sets and even all continuous images of separable complete metric spaces have the Baire property.
More info:
1. Meager is just another word for first category, i.e. the countable union of nowhere dense sets.
2. A set A in a topological space has the Baire property if for some open set V (possibly empty) the set (A-V)U(V-A) is meager.
3. The collection of sets with the Baire property form a sigma-algebra. All open sets trivially have the Baire property, thus all Borel sets have the Baire property. All analytic sets also have the Baire property.
4. Pettis Lemma: Let G be a topological group and let A be a non-meager subset of G with the Baire property. Then the set A*A^{-1} (element-wise multiplication) contains an open neighborhood of the identity. This is an analog to a similar theorem about Lebesgue measure: If A is a Lebesgue measurable subset of the reals with positive Lebesgue measure, then A - A (element-wise subtraction) contains an open set around 0.
 A: EDIT:  The following argument is in error.  See Tony Prochazka's answer.
I think you can have such a subspace. Let $f : X \to R$ be a discontinous linear functional (such a functional exists assuming Axiom of Choice, see wikipedia). The claim is that its kernel $K = \ker f$ is a proper non-meager subspace. It is definitely proper. Assume it would be meager. Then it is contained in the countable union of closed subsets $A_i$. Since $K=\ker f$ it has codimension 1, so there is $z \in X$ such that every $x \in X$ can be written as $x = k + az$, for some number $a$ and some $k \in K$. Let $B_i$ be the set of elements $A_i + [-i,i]z$ (that is the set of $x \in X$ for which $x = k+az$ where $k$ is in $A_i$ and $a$ in $[-i,i]$). Then, I think, $B_i$ are closed and they have to have empty interior. Indeed, if there is a small ball around $k + az$ in $B_i$ then $f$ will be continuous at $k+az$, and then (since $f$ is linear) continuous everywhere, contradicting the choice of $f$. Thus $B_i$ are closed and nowhere dense, but their union is then the whole space $X$, which contradicts Baire's theorem.
A: This is more of a question than an answer, but hopefully it helps.  What happens if Y is the kernel of a discontinuous linear functional on X?  Such functionals are easy to "construct" using Zorn's Lemma for the existence of a linear basis for X (X infinite dimensional of course).  In that case, Y is not closed and has codimension 1, so it is dense.  It seems to me that it would be non-meager, but I don't have an argument for why it is non-meager. 
A: I am afraid that Konstantin's accepted answer is seriously flawed.
In fact, what seems to be proved in his answer is that $\ker f$ is of second category, whenever $f$ is a discontinuous linear functional on a Banach space $X$. This assertion has been known as Wilansky-Klee conjecture
and has been disproved by Arias de Reyna under Martin's axiom (MA). 
He has proved that, under (MA), in any separable Banach space there exists a discontinuous linear functional $f$ such that $\ker f$ is of first category.
There have been some subsequent generalizations, see Kakol et al.
So, where is the gap in the above proof?
It is implicitly assumed that $\ker f = \bigcup A_i$. Then $f$ is bounded on $B_i=A_i+[-i,i]z$. But in reality, we have only $\ker f \subset \bigcup A_i$ and we cannot conclude that $f$ is bounded on $B_i$.
And finally, what is the answer to the OP's question?
It should not be surprising (remember the conjecture of Klee and Wilansky) that the answer is:
in every infinite dimensional Banach space $X$ there exists a discontinuous linear form $f$ such that $\ker f$ is of second category.
Indeed, let $(e_\gamma)_{\gamma \in \Gamma}$ be a normalized Hamel basis
 of $X$. 
Let us split $\Gamma$ into countably many pairwise disjoint sets $\Gamma =\bigcup_{n=1}^\infty \Gamma_n$, each of them infinite. We put $X_n=span\{e_\gamma: \gamma \in \bigcup_{i=1}^n \Gamma_i\}$. It is clear (from the definition of Hamel basis) that $X=\bigcup X_n$. Therefore there exists $n$ such that $X_n$ is of second category. Finally, we define $f(e_\gamma)=0$ for every $\gamma \in \bigcup_{i=1}^n\Gamma_i$ and $f(e_{\gamma_k})=k$ for some sequence $(\gamma_k) \subset \Gamma_{n+1}$. We extend $f$ to be a linear functional on $X$. It is clearly unbounded, $f\neq 0$, and $X_n \subset \ker f$. Hence $\ker f\neq X$ is dense in $X$ and of second category in $X$.
A: By subspace, do you mean "linear subspace"? Because if not, I don't see what's wrong with taking X to be the real line and Y to be the rationals.
