Is a non-meagre set comeagre in its closure?

Question

Let $A$ be a non-meagre subset of a topological space $X$. Is $A$ comeagre in its closure $\mathcal{Cl}(A)$?

Answer 1

Not necessarily. Let $A$ be a Bernstein set in $\Bbb R$. $A$ and $\Bbb R\setminus A$ are dense in $\Bbb R$ and non-meagre. In fact, both are Baire spaces.

Answer 2

No. Work in $\mathbb R$, and let $X$ be $\left((0,1)\cap\mathbb Q\right)\cup(1,2)$, so $X$ is nonmeagre (because of its right half) but non-co-meagre in its closure (because of its left half).