$\newcommand{\cl}{\operatorname{cl}}$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.)
Best Answer
Yes, an $F_\sigma$ set of the first category is the union of countably many closed nowhere dense sets.
Since an $F_\sigma$ set of the first category is the union of countably many closed sets of the first category, it will suffice to show that a closed set of the first category is the union of countably many closed nowhere dense sets. Well, a closed set of the first category is the union of countably many nowhere dense sets, and it is also the union of the closures of those nowhere dense sets, and the closure of a nowhere dense set is a closed nowhere dense set, Q.E.D.
Actually, a closed set of the first category (in a Baire space) is nowhere dense; so there is a shorter proof, if you're working with a Baire space.