Here is one example. Let $(q_n)_n$ be a list containing all rationals in $[0,1]$, let $f_n:[0,1]\to\mathbb{R}$ be given by $f_n(x)=1$ if $|x-q_n|<\frac{1}{n!}$ and $f_n(x)=0$ if not. Let $f=\sum_nf_n:[0,1]\to\mathbb{R}$.

If some function $g$ is in $[f]$ then it has to be unbounded in any open interval, so it cannot be continuous at any point.

Here is one example. Let $(q_n)_n$ be a list containing all rationals in $[0,1]$, let $f_n:[0,1]\to\mathbb{R}$ be given by $f_n(x)=1$ if $|x-q_n|<\frac{1}{n!}$ and $f_n(x)=0$ if not. Let $f=\sum_nf_n:[0,1]\to\mathbb{R}$.

If some function $g$ is in $[f]$ then it has to be unbounded in any open interval $(a,b)$ (so it cannot be continuous at any point): this is because inside any interval $(a,b)$ we can find by recursion an increasing sequence $(n_k)_k$ such that $q_{n_k}\in(a,b)$ and $|q_{n_{k+1}}-q_{n_k}|<\frac{1}{2n_k!}$. Thus, for all $k\in\mathbb{N}$, $f(x)\geq k$ for all $x$ in some nhood of $q_{n_k}$.