1.13

Part a.) Let $F \subset \mathbb{R}^{d}$ be an arbitrary closed set. First, define: $$O_n = \lbrace x \in \mathbb{R} \hspace{0.25cm} | \hspace{0.25cm} \mathbb{d}{(x,F)} < \frac{1}{n} \rbrace$$ Notice $\overline{F} = \bigcap_{n = 1}^{\infty} O_n$. Since $F$ is closed, $F = \overline{F}$. Thus $F \in F_\sigma$.

Part b.) There's a well-known corollary to Baire's Category Theorem that states:

$\dagger$ In a complete metric space, the intersection of any countable collection of dense $G_\delta$'s is again a dense $G_\delta$.

First notice that we can write $\mathbb{Q} = \bigcup_{n=1}^{\infty} \lbrace q_n \rbrace$, where $\lbrace q_n \rbrace_{n \in \mathbb{N}}$ is an enumeration of the rationals. Since each $\lbrace q_n \rbrace$ is closed, it's easy to see that $\mathbb{Q} \in F_\sigma$.

Next, observe that $\mathbb{Q} \in F_\sigma \Rightarrow \mathbb{Q}^c \in G_\delta$. Now it is clear from $\dagger$ that $\mathbb{Q}$ can not be $G_\delta$, since $\mathbb{Q} \cap \mathbb{Q}^c = \emptyset$, which is nowhere-dense. Thus, $\mathbb{Q}$ is $F_\sigma$ but not $G_\delta$, as desired.

Part c.) $$S = (\mathbb{Q} \cap (-\infty, 0]) \cup (\mathbb{Q}^c \cap (0, \infty))$$