Let $$\begin{split}\mathcal D(T_3)&:=\left\{f\in L^2([0,1];\mathbb C):f\text{ is absolutely continuous with }f'\in L^2([0,1];\mathbb C)\right\},\\\mathcal D(T_2)&:=\left\{f\in\mathcal D(T_3):f(0)=f(1)\right\},\\\mathcal D(T_1)&:=\left\{f\in\mathcal D(T_2):f(0)=0\right\},\\\mathcal D(T_0)&:=\mathcal D(T_1)\cap C^\infty([0,1];\mathbb C)\end{split}$$ and $$T_k:={\rm i}f'\;\;\;\text{for }\mathcal D(T_k)\text{ and }k\in\{0,\ldots,3\}.$$
What's the easiest way to observe that $T_0$ is closable and $\overline{T_0}=T_1$?
First of all, $T_0$ should be densely-defined, since $C([0,1];\mathbb C)$ is dense in $L^2([0,1];\mathbb C)$, right? Now, how do we observe that $T_1$ is closed? I was able to show that $T_1^\ast=T_3$ which yields that $T_3$ is closed (since the adjoint is always closed).
For simplicity, forget about the imaginary unit $i$, we are to show that, for $f_{n}\rightarrow f$ in $L^{2}$, and $f_{n}'\rightarrow g$ in $L^{2}$, it will follow that $g=f'$ a.e.
Since $\|\cdot\|_{L^{1}[0,1]}\leq\|\cdot\|_{L^{2}[0,1]}$, it is easy to see that \begin{align*} \int_{0}^{x}g(t)dt=\lim_{n\rightarrow\infty}\int_{0}^{x}f_{n}'(t)dt=\lim_{n\rightarrow\infty}f_{n}(x) \end{align*} But $f_{n}\rightarrow f$ in $L^{2}$ entails an almost everywhere pointwise convergent subsequence $f_{n_{k}}(x)\rightarrow f(x)$, this leads to \begin{align*} \int_{0}^{x}g(t)dt=\lim_{k\rightarrow\infty}f_{n_{k}}(x)=f(x)~~~~\text{a.e.} \end{align*} We are done.