Let $M=\{f(x) \in C[0,1]\mid f(0)=0\}$. Is $\overline {M}=L^2[0,1]$? The closure is taken under the usual norm of $L^2[0,1]$.
Moreover,fix n different points$(a_k)\subseteq[0,1]$,k=1...n, Let$\widetilde M=\{f(x) \in C[0,1]\mid f(a_k)=0,k=1...n\}$, is $\overline{\widetilde M}=L^2[0,1]$?
I'm assuming you know $C([0,1])$ is dense in $L^2([0,1])$, so it suffices to show the closure of $M$ in $L^2([0,1])$ contains $C([0,1])$.
Let $f\in C([0,1])$. let $g_n=\min(nx,1)$. Then $g_nf\in M$. Moreover, $fg_n$ is uniformly bounded and converges pointwise to $1_{x>0}f(x)$. Hence by the bounded convergence theorem,
$$\int_0^1 (f-(g_nf))^2\to0$$
as $n\to\infty$, i.e., $g_nf\to f$ in $L^2([0,1])$