fa.functional analysis - Are proper linear subspaces of Banach spaces always meager?

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 $gamma in bigcup_{i=1}^nGamma_i$ 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, $fneq 0$, and $X_n subset ker f$. Hence $ker fneq X$ is dense in $X$ and of second category in $X$.