$C([0,1])$ is not a closed subset of $L^1([0,1])$ with $L^1$ norm.
Does this counter example work ?
$f_n(x)= \begin{cases} n &x\in(0,\frac{1}{n^2}) \\ \frac{1}{\sqrt{x}} &x\in (\frac{1}{n^2},1) \\ \end{cases} $
This converges to $\frac{1}{\sqrt{x}}$, This function is Lebesgue integrable but not continuous.
Can I safely say that $C^n[0,1]$ for $0\leq n \leq \infty$ is not closed in $L^1[0,1]$ ?
Yes, it works. You may want to use $f(x)=\frac{1}{\sqrt{\left\lvert x-\frac12\right\rvert}}$, but it is essentially the same. An analytic example is $f_n(x)=\frac{(2x)^n}{1+(2x)^n}$, which is dominated by the constant $1$ and converges pointwise (and thus in $L^1$) to the integrable function $$f(x)=\begin{cases}0&\text{if }x<\frac12\\ \frac12&\text{if }x=\frac12\\ 1&\text{if }x>\frac12\end{cases}.$$
The latter is not almost-everywhere equal to any continuous function.