 Solution

A subset is dense when every nonempty open set meets it, equivalently its closure is the whole space. Baire category theorem says a complete metric space has dense intersection for every countable family of open dense sets. Hereso Baire makes $S$ dense. It need not be open: enumerate the rationals $q_j$ and take $A_j=[0,1]\setminus \{q_1,\dots ,q_j\}$, giving the irrationals. Nor need ${\cap }_j{A}_j$ be nonempty: take $A_1=\varnothing$ and $A_j=[0,1]$ for $j\ge 2$.