Consider $A:\mathcal{D}(A)\subset L^{2}[0,1]\to L^{2}[0,1]$ given by $$A:=-\frac{d^{2}}{dx^{2}},\qquad\mathcal{D}(A):=C^{2}_{0}(0,1)$$
Now, I assume that the closure of $A$ is its extension defined on $C^{2}_{0}[0,1]$. Then am I right in thinking that this is this densely defined if $\overline{C_{0}^{2}[0,1]}=L^{2}[0,1]$?
Your operator is densely-defined, and it is closable. So the closure of the operator in question is densely-defined as well.
I'm not sure about your notation $\mathcal{C}_0^{\infty}$; does it mean the functions vanish at the endpoints, or does it mean compactly supported in $(0,1)$? Unfortunately, I've seen that notation used in both ways.
If $C_0^{\infty}(0,1)$ is the initial domain $\mathcal{D}(A)$, and we're assuming compact support, then the closure of $A$ has a domain consisting of $f \in L^2[0,1]$ which may be modified on a set of measure $0$ to obtain a twice absolutely continuous function $\tilde{f}$ on $[0,1]$ such that $\tilde{f}(0)=\tilde{f}'(0)=\tilde{f}(1)=\tilde{f}'(1)=0$, and $\tilde{f}''\in L^2$.