[proofplan]
For each ground-model closed [nowhere dense set](/page/Nowhere%20Dense%20Set) $F_n$, we define a dense set of Cohen conditions whose basic clopen neighbourhoods avoid $F_n$. Cohen genericity guarantees that the generic filter determined by $c$ meets each of these dense sets. Hence, for every $n$, some initial segment of $c$ already forces $c$ into a basic neighbourhood disjoint from $F_n$, so $c \notin F_n$. Since $A$ is contained in the union of the $F_n$, this gives $c \notin A$.
[/proofplan]
[step:Build the dense set of conditions avoiding one closed nowhere dense set]
Fix $n \in \mathbb{N}$. Define
\begin{align*}
D_n := \{s \in 2^{<\mathbb{N}} : [s] \cap F_n = \varnothing\}.
\end{align*}
We claim that $D_n$ is dense in $\mathbb{C}$.
Let $t \in 2^{<\mathbb{N}}$. Let $\overline{F_n}$ denote the closure of $F_n$ in the topology of $2^{\mathbb{N}}$. Since $F_n$ is nowhere dense in $2^{\mathbb{N}}$, the non-empty basic [open set](/page/Open%20Set) $[t]$ is not contained in $\overline{F_n}$. Since $F_n$ is closed, $\overline{F_n} = F_n$, so there exists $x \in [t] \setminus F_n$ and an open neighbourhood $U \subset 2^{\mathbb{N}}$ of $x$ such that $U \cap F_n = \varnothing$. The basic clopen sets form a base for Cantor space, so choose $s \in 2^{<\mathbb{N}}$ with $t \subset s$, $x \in [s]$, and $[s] \subset U$. Then $s \leq t$ in $\mathbb{C}$ and $[s] \cap F_n = \varnothing$, so $s \in D_n$. Thus $D_n$ is dense.
Moreover, because the code for $F_n$ belongs to $M$ and the definition of $D_n$ is first-order over the coded [closed set](/page/Closed%20Set), we have $D_n \in M$.
[guided]
Fix one index $n \in \mathbb{N}$. Our goal is to find many finite binary strings whose cylinders avoid $F_n$. Define
\begin{align*}
D_n := \{s \in 2^{<\mathbb{N}} : [s] \cap F_n = \varnothing\}.
\end{align*}
This is the set of Cohen conditions that already decide the generic real to lie outside $F_n$, because any real extending such an $s$ belongs to $[s]$, and $[s]$ is disjoint from $F_n$.
We verify that $D_n$ is dense in $\mathbb{C}$. Let $t \in 2^{<\mathbb{N}}$ be an arbitrary Cohen condition. To prove density, we must find an extension $s \leq t$, which means $t \subset s$, such that $s \in D_n$.
The basic open set determined by $t$ is
\begin{align*}
[t] := \{x \in 2^{\mathbb{N}} : t \subset x\}.
\end{align*}
It is non-empty and open in Cantor space. Let $\overline{F_n}$ denote the closure of $F_n$ in the topology of $2^{\mathbb{N}}$. Since $F_n$ is nowhere dense, no non-empty open set is contained in $\overline{F_n}$. Therefore $[t]$ is not contained in $\overline{F_n}$. Because $F_n$ is closed, $\overline{F_n} = F_n$, and hence there exists some $x \in [t]$ with $x \notin F_n$.
Now we need more than a single point outside $F_n$; we need an entire basic neighbourhood outside $F_n$. Since $F_n$ is closed, its complement $2^{\mathbb{N}} \setminus F_n$ is open. The point $x$ belongs to this complement, so there is an open set $U \subset 2^{\mathbb{N}}$ such that $x \in U$ and $U \cap F_n = \varnothing$. The basic clopen cylinders form a base for the topology of $2^{\mathbb{N}}$, so there exists a finite binary string $s \in 2^{<\mathbb{N}}$ such that $x \in [s] \subset U$. Since $x \in [t]$ as well, we may choose $s$ long enough so that $t \subset s$. Then $s \leq t$ in the Cohen order and
\begin{align*}
[s] \cap F_n = \varnothing.
\end{align*}
Thus $s \in D_n$. Since every $t \in 2^{<\mathbb{N}}$ has such an extension, $D_n$ is dense in $\mathbb{C}$.
Finally, $D_n$ belongs to $M$. The set $F_n$ is coded in $M$, the forcing poset $2^{<\mathbb{N}}$ belongs to $M$, and the relation $[s] \cap F_n = \varnothing$ is definable from the code for $F_n$. Therefore the model $M$ contains the dense set $D_n$.
[/guided]
[/step]
[step:Use Cohen genericity to avoid each closed nowhere dense set]
Since $D_n \in M$ is dense in $\mathbb{C}$ and $c$ is Cohen generic over $M$, the filter
\begin{align*}
G_c := \{s \in 2^{<\mathbb{N}} : s \subset c\}
\end{align*}
meets $D_n$. Hence there exists $s_n \in 2^{<\mathbb{N}}$ such that $s_n \subset c$ and $[s_n] \cap F_n = \varnothing$. Since $c \in [s_n]$, it follows that $c \notin F_n$.
Because $n \in \mathbb{N}$ was arbitrary, we have
\begin{align*}
c \notin F_n
\end{align*}
for every $n \in \mathbb{N}$.
[/step]
[step:Conclude that the generic real is not in the meagre set]
The hypothesis gives
\begin{align*}
A \subseteq \bigcup_{n \in \mathbb{N}} F_n.
\end{align*}
If $c \in A$, then $c \in F_n$ for some $n \in \mathbb{N}$, contradicting the conclusion of the previous step. Therefore $c \notin A$, as required.
[/step]
