Taking two times the sequential closure of the set of continuous functions in the topology of pointwise convergence? Consider the unit interval $I=[0, 1]$ and assume that the function $f\colon I\to \mathbb R$ satisfies 
$$
f(t)=\lim_{n\to \infty} f_n(t), \qquad \text{for all }t\in I
$$
where 
$$
f_n(t)=\lim_{j\to \infty} f_{n\, j}(t), \qquad \text{for all }t\in I
$$
and $f_{n\, j}\in C(I)$. 

Question. Is it true that there exists a sequence $g_n\in C(I)$ such that $$f(t)=\lim_{n\to \infty} g_n(t), \qquad \text{for all }t\in I?$$

The answer is positive if every "$\text{for all}$" is replaced by "$\text{for almost all}$", and the proof uses in an essential way Egorov's theorem to obtain some uniformity. (See Proposition 1.4.3 here). That's why I think that the answer to the present question is negative.
 A: The answer is negative. In fact if we define the Baire class $B_\alpha$ for countable ordinals $\alpha$ by saying $B_0=C(I)$, $B_{\alpha+1}$ is the set of pointwise limits of sequences in $B_\alpha$, and $B_\alpha=\bigcup_{\beta<\alpha}B_\beta$ for limit ordinals $\alpha$ then all the $B_\alpha$ are distinct. 
Or so I've read; don't ask me to prove it. Your question asks whether $B_2=B_1$, and it's easy to give an example showing that that's not so.
Say $S=[0,1]\cap\Bbb Q$ and let $f=\chi_S$. It's clear that the characteristic function of any finite set is in $B_1$, and hence $f\in B_2$. It follows from the Baire Category Theorem that $f\notin B_1$:
If $(f_n)$ is a sequence of continuous functions then $\{x:\limsup f_n(x)>1/2\}$ is a $G_\delta$. But $S$ is not a $G_\delta$. Say $S=\{r_1,r_2,\dots\}$, and let $S_n=\{r_n,r_{n+1}\dots\}$. If $S$ were a $G_\delta$ then each $S_n$ would be a dense $G_\delta$ and hence $\emptyset$ would be dense.
A: Ciao Giuseppe.
The answer is negative, as @DavidC.Ullrich has already pointed out. I can provide an explicit counterexample. Consider the function $q:\mathbb{Q}\backslash\{0\}\to\mathbb{N}$ associating to $x$ the unique natural number $q=q(x)$ such that we can write $x=\tfrac{p}{q}$ an irreducible fraction. Then define
$$f(x) = \cases{q(x)&if $x\in\mathbb{Q}\backslash\{0\}$,\\0&else.}$$
This function is not the pointwise limit of any sequence of continuous functions (see e.g. this answer to a question about pointwise limits of continuous functions). Now it is easy to construct sequences as in your question. let $\{x_k\}_k$ be an enumeration of all non-zero rationals in $I$. For any number $x\in I$, let $\{g_j^{x}\}_j$ be a sequence of continuous functions converging pointwise to $\chi_{\{x\}}$ (easily constructed). Now take
$$f_{n,j} = \sum_{k=1}^nq(x_k)g_j^{x_k}.$$
Then
$$f_{n,j}\stackrel{j\to\infty}{\longrightarrow}\sum_{k=1}^nq(x_k)\chi_{\{x_k\}}=:f_n$$
pointwise, and
$$f_n\stackrel{n\to\infty}{\longrightarrow}f$$
pointwise.
