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 $X$.
On the relation between null sets and meagre sets, you can also look at
this article.
Two theorems mentioned in this note (both classical and not due to the author):
(As already mentioned above) There exist a meagre $F_\sigma$ subset $A$ and a null $G_\delta$ subset $B$ of $\mathbb R$ that satisfy $A\cap B=\emptyset$ and $A\cup B=\mathbb R$.
(The Erdős-Sierpiński Duality Theorem) Assume that the Continuum Hypothesis
holds. Then there exists an involution (bijection of order two) $f:\mathbb R\to\mathbb R$ such that $f[A]$
is meagre if and only if $A$ is null, and $f[A]$ is null if and only if $A$ is meagre for every subset $A$ of $\mathbb R$.
While (1) says that the ideals of null, respectively meager sets are "orthogonal", (2) says that assuming CH they behave identically. But it is well known that this duality between measure and category fails dramatically once we take a more abstract point of view: Shelah proved that you need large cardinals to construct a model of set theory (ZF, no axiom of choice) where every set of reals is Lebesgue measurable, but no large cardinals are necessary to construct a model where every set of reals has the Baire property (the corresponding notion to measurability for category).
Best Answer
First, consider Gerhard's easier special case, where the open sets are disjoint.
Claim. The union of an arbitrary family of pairwise disjoint open meager sets is meager.
Proof. Suppose that $U_i$ are pairwise disjoint and meager, so that $U_i\subset\bigcup_n C_n^i$, where each $C_n^i$ is closed and nowhere dense. Let $C_n=\bigcup_i (C_n^i\cap U_i)$. This set is not dense on any nonempty open set, because if it were dense on some nonempty $V$, then it would be dense on some nonempty $V\cap U_i$, but that is impossible since by the disjointness hypothesis only $C_n^i$ contributes points to this set, and it is nowhere dense. Thus, the closure of $C_n$ is closed and nowhere dense, and $\bigcup_i U_i$ is contained within $\bigcup_n C_n$, since each $U_i$ is contained within and is in fact equal to $\bigcup_n (C_n^i\cap U_i)$. So $\bigcup_i U_i$ is meager. QED
A similar idea works in general, by well-ordering the family of open sets.
Theorem. An arbitrary union of open meager sets is meager.
Proof. Suppose we have a family of open meager sets $U_\alpha$, indexed by ordinals $\alpha$, so that for each $\alpha$ we have $U_\alpha\subset\bigcup_n C_\alpha^n$ for some closed nowhere dense sets $C_\alpha^n$. Let $$C_n=\bigcup_\alpha [C_\alpha^n\cap U_\alpha-\bigcup_{\beta\lt\alpha}U_\beta].$$ Note that these $C_n$ cover the union $U=\bigcup_\alpha U_\alpha$, since any $a\in U$ is in some least $U_\alpha$ and so it gets into some $C_\alpha^n\cap U_\alpha$ without being in $\bigcup_{\beta\lt\alpha}U_\beta$, and consequently is in $C_n$. Also, each $C_n$ is nowhere dense, because if $C_n$ is dense on some nonempty set $V$, then there is some least $\alpha$ containing members of $V$, and so we reduce to nonempty $V\subset U_\alpha-\bigcup_{\beta\lt\alpha}U_\beta$; thus, $C_n$ would be dense on $V$ inside $U_\alpha-\bigcup_{\beta\lt\alpha}U_\alpha$. But the only members of this set in $C_n$ are contributed by $C_\alpha^n$, which is nowhere dense. So the closure of $C_n$ is nowhere dense, and so $U$ is meager, as desired. QED