Baire's theorem: category and density for complements of first category sets

Question

I'm working with the following version of Baire's category theorem:

If a non-empty complete metric space $(M,d)$ is the countable union of closed sets, then one of these closed sets has non-empty interior.

I want to show that if $A\subset M$ is a set of first category then $A^c := M\setminus A$ is a set of second category and dense in $M$.

The equivalent versions of Baire's theorem have me confused as I am very new to the concept of Baire categories. I tried working with the following statement:

$A$ is a set of first category (i.e. $A = \bigcup_{n \in \mathbb{N}} A_n$ and for all $n$ holds $A_n$ is nowhere dense) iff for all $n$ the set $(\overline{A_n})^c$ is dense in $M$.

The obvious proof by taking $A$ to the complement needs to assume that in a complete metric space the intersection of countably many dense open sets is dense. I read that this is the implication of Baire's lemma, so I guess I cannot just assume this holds true. The necessary step should relate to the statement of the theorem, however, even after reading the referenced post, I do not see how this is in accordance with this version of it.

Answer 1

Assume $A$ is of first category, and assume, toward a contradiction, that its complement $A^c$ is also of first category. So $A$ is the union of countably many nowhere dense sets $A_n$ and $A^c$ is the union of countably many nowhere dense sets $B_n$. By definition of "nowhere dense", the closures $\overline{A_n}$ and $\overline{B_n}$ have empty interiors. But the union of all these closures inlcudes both $A$ and $A^c$, so it is the whole space, and that contradicts Baire's theorem.