The recipe for these kind of problems is to write down the formula for a point $x$ to be in this set, using countable sets (like $\frac{1}{n}, n \in \mathbb{N}$, instead
of arbitrary $\epsilon$ and $\delta$). Also use that $f_n(x)$ converges iff it is a Cauchy sequence in $\mathbb{R}$, as $\mathbb{R}$ is complete.
So $x$ is in this set iff
for all $n$ in $\mathbb{N}$ there exists $m$ in $\mathbb{N}$ such that $k,l \ge m$ implies that $|f_k(x) - f_l(x)| \le \frac{1}{n}$.
Now define $A_{k,l,n} = \left\{x: | f_k(x) - f_l(x) | \le \frac{1}{n} \right\}$ which is closed (as the inverse image of the continuous function $|f_k - f_l|$ of the set $[0,\frac{1}{n}]$, eg.).
So the set of convergence points of $(f_n)$ equals $\cap_n \cup_m \cap_{k,l \ge m} A_{k,l,n}$
where the last set is closed, as an intersection of closed sets, and so
this set is a countable intersection of a countable union of closed sets.
As it was mentioned here, the statement in question was proved by Hahn in 1919 and independently by Sierpinski in 1921. The proof can be found also in Hausdorff's book Set Theory and in Kechris' Classical Descriptive Set Theory. The most accessible proof is that from Kechris' book, and is essentially the same as Hausdorff's one. Original Hahn's proof uses a similar argument, but is slightly more difficult to follow. Sierpinski's proof is based on a completely different idea and I had a problem to understand it. You can find its outline in the above referenced discussion. I will sketch the Kechris' version of the proof here, for those who do not want to look for it in the book.
An important tool used in the proof is the following lemma due to Baire.
Let $f:\mathbb{R}\to [0,\infty]$ be lower semicontinuous, i.e., preimages of $(a,\infty]$ are open.
Then there exists a nondecreasing sequence $\langle f_n:n\in\mathbb{N}\rangle$ of continuous functions converging to $f$.
The most elegant way to prove the lemma is to use Yosida regularization, namely, to define $f_n(x)=\inf\{f(y)+n\lvert x-y\rvert:y\in\mathbb{R}\}$.
Then we can prove a special case of the theorem.
Let $E\subseteq\mathbb{R}$ be $F_\sigma$. Then there exists a sequence $\langle g_n:n\in\mathbb{N}\rangle$ of continuous functions $g_n\colon\mathbb{R}\to [-1,1]$ such that $g_n(x)\to 0$ if $x\in E$, and $f_n(x)$ diverges if $x\notin E$.
Proof. Assume that $E=\bigcup_n E_n$, where $E_n$ is closed and $E_n\subseteq E_{n+1}$, for every $n$. Then the function $f:\mathbb{R}\to[0,\infty]$ defined by $f(x)=\min\{n:x\in E_n\}$ if $x\in E$, and $f(x)=\infty$ for $x\notin E$, is lower semicontinuous. By the lemma, there exists a nondecreasing sequence $\langle f_n:n\in\mathbb{N}\rangle$ of continuous functions converging to $f$. We may also assume that for every $x$, $\lvert f_{n+1}(x)-f_n(x)\rvert\le 1/2$. This can be achieved by taking $f'_n(x)=\min\{n,f_n(x)\}$ and then inserting some terms into the sequence $\langle f'_n:n\in\mathbb{N}\rangle$. Finally, we put $g_n(x)=\sin(\pi f_n(x))$. If $x\in E$ then $f_n(x)$ converges to some integer and hence $g_n(x)$ converges to $0$. If $x\notin E$ then $f_n(x)$ converges to $\infty$, and since $\lvert f_{n+1}(x)-f_n(x)\rvert\le 1/2$, it is clear that $g_n(x)$ oscillates. This finishes the proof of the special case.
To finish the proof of the theorem, assume that $E=\bigcap_k E_k$, $E_k$ being $F_\sigma$. For every $k$, there exists a sequence $\langle g^k_n:n\in\mathbb{N}\rangle$ of continuous functions, $g^k_n\colon\mathbb{R}\to [-1,1]$, such that $g^k_n(x)\to 0$ if $x\in E_k$, and $g^k_n(x)$ diverges if $x\notin E_k$. Let us take a sequence consisting of all functions $2^{-k}g^k_n$, for $k,n\in\mathbb{N}$. As it can be easily checked, the obtained sequence has the desired properties.
It seems that it would be quite difficult to rewrite this proof into a direct definition of the sequence $\langle f_n:n\in\mathbb{N}\rangle$ from the sets $E^k_n$ mentioned in the question. Also, in my unsuccessful attempts I was trying to use only the normality of $\mathbb{R}$ and the fact that open sets of reals are $F_\sigma$. However, the proof of the Baire's statement depend heavily on the metric, so I was trying probably to prove also a kind of metrization theorem. It would be interesting to know if the result holds also in non-metrizable spaces.
Best Answer
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 $k \in \Bbb N$.