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- Every open set is union of balls with rational radius and rational center.
- Every open ball is a countable union of closed balls.
This gives (ii). For (i), given two points in your open set, say that they are equivalent iff there is a continuous path between them, completely contained in the open set. Argue that this is indeed an equivalence relation, and that its components (equivalence classes) are open. Now use that $\mathbb Q^n$ is dense in $\mathbb R^n$, so there can be no more than countably many equivalence classes.
Actually I feel a hint is not enough here. I needed to think a bit more as well so here is an attempt at a solution.
The first identity to notice is this:
$$C = \bigcap_{k \in \Bbb N} \; \underbrace {\{ x \in \Bbb R\ : \exists N \in \Bbb N \; \text{s.t.} \; m, n \gt N \implies |f_m(x) - f_n(x)| \le \frac 1 k \}}_{B_k} $$
The second is:
$$B_k = \bigcup_{n \in \Bbb N} \; \underbrace{\{ \ x \in \Bbb R : |f_m(x) - f_n(x)| \le \frac 1 k\ \; \text{for each} \;m\ge n \}}_{A_{n, k}}$$
The third is:
$$A_{n, k} = \bigcap_{t \ge n} \; \underbrace{\{x \in \Bbb R : |f_t(x) - f_n(x)| \le \frac 1 k \}}_{F_{t, n,k}}$$
Now each $F_{t, n,k}$ is a closed set since it is the pre-image of a closed set under a continuous function, $g = |f_t - f_n|$. Since any intersection of closed sets is closed, $A_{n,k}$ is closed. Hence, $B_k$ is an $F_\sigma$ set for each $k \in \Bbb N$.
Best Answer
Summarising comments, essentially:
You don't need to use $F_\sigma$ sets at all. It should be well-known that $\Bbb N \times \Bbb N$ is countable and so if $G_n = \bigcap_{m \in \Bbb N} O^{(n)}_m$ is a sequence of $G_\delta$'s in a space $X$ (so all $O^{(n)}_m$ are open sets) then $$\bigcap_{n \in \Bbb N} G_n = \bigcap_{(n,m) \in \Bbb N \times \Bbb N} O^{(n)}_m$$ is also a countable intersection of open sets, and hence a $G_\delta$.
But if $F_\sigma$ are absolutely necessary, apply de Morgan twice: $$\left(\bigcap_n \bigcap_m O^{(n)}_m\right)^\complement =\bigcup_n \left(\bigcap_m O^{(n)}_m \right)^\complement = \bigcup_n \bigcup_m \left(O^{(n)}_m\right)^\complement $$ so the complement of $\bigcap_n G_n$ is an $F_\sigma$ (a countable union of $F_\sigma$ sets was supposed to be known) and so the set itself is $G_\delta$. But IMHO this is too roundabout a way.