If $\{X_t\}_{t\ge 0}$ is a simple process. i.e.$0=t_0\le t_1\le\cdots\le t_n=T$
$\exists \xi_i\in\mathcal F_{t_i}$ s.t.$X_t(\omega)=\xi_i(\omega)$ when $t\in[t_i,t_{i+1}].$
$\{W_t\}_{t\ge 0}$ is a B.M. So we have: $$\mathbb E\left[(\int_0^T X_t\text{d}W_t)^2\right]=\mathbb E\left[\int_0^TX_t^2\text{d}t\right]$$
I want to prove it: $$(\int_0^T X_t\text{d}W_t)^2=(\sum_{i=0}^{n-1}\xi_i(W_{i+1}-W_i))^2=\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\xi_i\xi_j(W_{i+1}-W_i)(W_{j+1}-W_j)$$
if $i\neq j$,$(W_{i+1}-W_i)$and$(W_{j+1}-W_j)$ are independent.
If $$\mathbb E\left[\xi_i\xi_j(W_{i+1}-W_i)(W_{j+1}-W_j)\right]=\mathbb E\left[\xi_i\xi_j\right]\mathbb E\left[(W_{i+1}-W_i)(W_{j+1}-W_j)\right]$$ then I can finish the proof,how to obtain this equation ?(there is no information about independence)
Let $i < j$. Note that $$ \begin{split} E\left[\xi_i\xi_j(W_{i+1}-W_i)(W_{j+1}-W_j)\right] &= E\left[ E\left[\xi_i\xi_j(W_{i+1}-W_i)(W_{j+1}-W_j)|\mathcal F_{t_j}\right] \right] \\ &= E\left[\xi_i\xi_j(W_{i+1}-W_i) E\left[(W_{j+1}-W_j)|\mathcal F_{t_j}\right] \right] \\ &= 0 \end{split} $$ By symmetry $i > j$ will be $0$ as well.
For $i=j$, $$ \begin{split} E\left[\xi_i^2(W_{i+1}-W_i)^2\right] &= E\left[E\left[\xi_i^2(W_{i+1}-W_i)^2|\mathcal F_{t_i}\right]\right] \\ &= E\left[\xi_i^2 E\left[(W_{i+1}-W_i)^2|\mathcal F_{t_i}\right]\right] \\ &= E\left[\xi_i^2 \right](t_{i+1} - t_i) \\ \end{split} $$
We have $$ \begin{split} E\left[\left(\int_0^T X_t\text{d}W_t\right)^2\right] &= \sum_{i=0}^{n-1}\sum_{j=0}^{n-1}E\left[\xi_i\xi_j(W_{i+1}-W_i)(W_{j+1}-W_j)\right] \\ &= \sum_{i=0}^{n-1} E\left[\xi_i^2 \right](t_{i+1} - t_i) \\ &= E\left[\sum_{i=0}^{n-1} \xi_i^2 (t_{i+1} - t_i)\right] \\ &= E\left[\int_0^T X_t^2 dt\right] \\ \end{split} $$