MA makes unions of fewer than continuum many meagre sets meagre

By AC choose increasing closed nowhere-dense covers $M_{\alpha }\subseteq {\bigcup }_n{F}_{\alpha n}$. Fix a countable base $(B_j)$ of rational intervals. A condition is $(m,F,w)$, where $m<\omega$, $F\subseteq \kappa$ is finite, and $w(n,j)$ for $n,j<m$ is a nonempty rational interval with closure inside $B_j$. A condition $(m^{\mathrm{\prime }},F^{\mathrm{\prime }},w^{\mathrm{\prime }})$ extends $(m,F,w)$ when $m^{\mathrm{\prime }}\ge m$, $F^{\mathrm{\prime }}\supseteq F$, it preserves old $w$, and every newly assigned cell $(n,j)\in [0,m^{\mathrm{\prime }}{)}^2\setminus [0,m{)}^2$ has closure disjoint from $F_{\alpha n}$ for all $\alpha \in F$, the old side set. This includes the new columns of old rows. The relation is transitive: cells added at the first extension avoid the original side set, and cells added later avoid the larger intermediate side set. Finite unions of closed nowhere-dense sets leave the required subintervals in every $B_j$.

Conditions with the same $m,w$ are compatible: take the union of their finite side sets without adding a new row. There are only countably many finite rational arrays $w$, so the order is ccc. For each $\alpha$, the set requiring $\alpha \in F$ is dense. For every $r$, the set requiring $m>r$ is dense by filling the finitely many new cells inside the complements prescribed in step 1.1. These are only $\kappa +{\mathrm{\aleph }}_0<\mathfrak{c}$ requirements, so MA supplies a filter $G$.

Coherence of the filter defines $I_{nj}=w(n,j)$ for every $n,j$. Put $V_n={\bigcup }_j{I}_{nj}$ and $W_r={\bigcup }_{n\ge r}{V}_n$. Each $V_n$, hence each $W_r$, is open dense because $I_{nj}\subseteq B_j$ exists for every basic interval. Fix $\alpha$ and a filter condition that first has $\alpha$ in its side set at length $m$. For every $n\ge m$ and every $j$, the cell $(n,j)$ is assigned only in an extension of that condition and its closure avoids $F_{\alpha n}$ by step 1.1, even if it is a new column of an earlier row. Thus $V_n\cap F_{\alpha n}=\mathrm{\varnothing }$. If $x\in M_{\alpha }$, choose $k$ with $x\in F_{\alpha k}$; since the cover is increasing, $x\mathrm{\notin }{V}_n$ for all $n\ge \max (m,k)$. Thus $x\mathrm{\notin }{W}_{\max (m,k)}$ and $x\mathrm{\notin }{\bigcap }_r{W}_r$. Therefore the union of the $M_{\alpha }$ is covered by the meagre set ${\bigcup }_r(\mathbb{R}\setminus W_r)$. Singletons are nowhere dense, proving the final clause. The simultaneous cover choice is the exact AC use; the defective published sigma-ideal proposition is not used.