= 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. Here
$$
S=\bigcap_{N\ge1}\bigcup_{j\ge N}A_j,
$$
so Baire makes $S$ dense. It need not be open: enumerate the rationals $q_j$ and take $A_j=[0,1]\setminus\{q_1,\ldots,q_j\}$, giving the irrationals. Nor need $\cap_jA_j$ be nonempty: take $A_1=\varnothing$ and $A_j=[0,1]$ for $j\ge2$.
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