$\newcommand{\ep}{\epsilon}\newcommand{\de}{\delta}$The answer is no. To show this, let us use a slightly modified suggestion in a comment by Mateusz Kwaśnicki, as follows. Let
\begin{equation}
f_n:=g_{a_n,1/n},
\end{equation}
where
\begin{equation}
g_{a,h}(x):=1(a<x\le a+h)\frac{x-a}h+1(x>a+h)
\end{equation}
and $(a_n)$ is any sequence uniformly distributed on $[1/4,3/4]$, so that
\begin{equation}
\frac1N\sum_1^N 1(a<a_n\le a+h)\to\frac h{3/4-1/4}=2h
\end{equation}
(as $N\to\infty$) for any $a$ and $h$ such that $1/4\le a\le a+h\le3/4$. Note that $f_n$ is continuous and $0\le f_n\le1$, for each $n$.

Take now any $x$ and $y$ such that $0\le x<y\le1$. Then
\begin{align*}
\sum_{n=1}^N |f_n(x)-f_n(y)|=\sum_{n=1}^N (f_n(y)-f_n(x))\le S_1+S_2+S_2,
\end{align*}
where
\begin{equation}
S_1:=\sum_{n=1}^N 1(a_n\le x\le a_n+1/n)=\sum_{n=1}^N 1(x-1/n\le a_n\le x)=o(N),
\end{equation}
since $(a_n)$ is uniformly distributed on $[1/4,3/4]$; similarly,
\begin{equation}
S_2:=\sum_{n=1}^N 1(a_n\le y\le a_n+1/n)=o(N);
\end{equation}
and
\begin{equation}
S_3:=\sum_{n=1}^N 1(x\le a_n, y\ge a_n+1/n)\le\sum_{n=1}^N 1(x\le a_n\le y)\lesssim 2N(y-x).
\end{equation}
Thus,
\begin{align*}
\limsup_N\frac1N\sum_{n=1}^N |f_n(x)-f_n(y)|\le2|y-x|
\end{align*}
for any $x,y$ in $[0,1]$.

On the other hand, the limit $f$ of any subsequence $f_{n_k}$ that converges in measure is of the form $f(x)=1(x>a)$ for some $a\in[1/4,3/4]$ and almost all $x\in[0,1]$. So, the limit cannot be continuous.