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