Baire's property iff first category has dense complement. Show that $(S, d)$ has Baire's property iff every set of first category has a dense complement. 

A set is of first category if it is a countable union of nowhere dense sets. First Category
Baire's Lemma: Let $(X, \rho)$ be a complete metric space and $\{U_n\}_{n=1}^{\infty}$ a sequence of open dense sets in $X$. Then the set $\bigcap_{n =1}^{\infty} U_n$ is also dense.
Note that A is nowhere dense iff $(\overline{A})^c$ is open and dense. Suppose $\forall A$, $A$ is first category. That is, $A$ is a countable union of nowhere dense sets. i.e., $A_i$ is nowhere dense. $\forall A, \; A = \bigcup_{n \in \mathbb{N}}A_i \implies (A)^c = \bigcap_{n \in \mathbb{N}}(A_{i})^{c}$. But we don't know if this is a countable dense set! We know that $(\overline{A}_i)^c$ would be, but not just $(A)^c$.

If $A$ is nowhere dense, what can we say about the compliment of $A$? 
Can anyone clear this up or provide advice?

More scratch work:
In my proposed proof, I said let all $A \subset S$ be of first category. i.e., each $A_i$ is nowhere dense, i.e, $(\overline{A}_i)^c$ is open and dense. So, $\forall A, (\overline{A}_i)^c = \bigcup (\overline{A}_i)^c \implies ((\overline{A}_i)^c)^c = \bigcap (((\overline{A}_i)^c)^c) = \bigcap (\overline{A}_i)$. Still though, what can I say about $\overline{A}_i$?
 A: $\newcommand{\cl}{\operatorname{cl}}$Suppose first that $\langle S,d\rangle$ has the property that the intersection of countably many dense open sets is dense in $S$, and let $A$ be a first category set in $S$; we want to show that $S\setminus A$ is dense in $S$. Since $A$ is first category, there are nowhere dense sets $A_k$ for $k\in\Bbb N$ such that $A=\bigcup_{k\in\Bbb N}A_k$. For $k\in\Bbb N$ let $U_k=S\setminus\cl A_k$; each $U_k$ is a dense open subset of $S$, so $\bigcap_{k\in\Bbb N}U_k$ is dense in $S$. But
$$\bigcap_{k\in\Bbb N}U_k=\bigcap_{k\in\Bbb N}(S\setminus\cl A_k)=S\setminus\bigcup_{k\in\Bbb N}\cl A_k\subseteq S\setminus\bigcup_{k\in\Bbb N}A_k=S\setminus A\;,$$
so $S\setminus A$ is also dense in $S$.
Now suppose that the complement of every first category subset of $S$ is dense in $S$, and let $\{U_k:k\in\Bbb N\}$ be a family of dense open subsets of $S$. For each $k\in\Bbb N$ let $A_k=S\setminus U_k$; $A_k$ is closed and nowhere dense in $S$, so $A=\bigcup_{k\in\Bbb N}A_k$ is first category in $S$. Finally,
$$\bigcap_{k\in\Bbb N}U_k=\bigcap_{k\in\Bbb N}(S\setminus A_k)=S\setminus\bigcup_{k\in\Bbb N}A_k=S\setminus A\;,$$
which is dense in $S$, as desired.
A good book with much information on such topics is John C. Oxtoby, Measure and Category: A Survey of the Analogies between Topological and Measure Spaces, 2nd edition. (The first edition is also good.)
A: A nowhere dense set has empty interior. Therefore, the complement of a nowhere dense set is dense. Hence, I think my original solution is fine for one direction. If anyone else has comments to add, please do so. 
