I consider $\phi\in H_{2}^{2}$ the set of progressively measurable process such that
$$ \lVert \phi\rVert_{H_{2}^{2}} = \mathbb{E}\left(\int_{0}^{\infty}\phi_{s}^{2}ds\right) <\infty $$
I would like to compute the variance of $I(\phi)_{t}$ which is the Ito integral (with respect to a Brownian motion) of $\phi$ defined as a limit of integral of simpler class of function in $H_{2}^{2}$ namely the class
$$ \mathcal{R}=\text{span}\{\psi(\omega) 1_{[t_1, t_2[}(t) : t_1<t_2\in[0,+\infty[,\psi\in L^2(\mathcal{F}_{t_1} \} $$
For a function $a\in\mathcal{R}$ the Ito integral at time $t$ is defined by $\sum_{i=0}^{n-1}\psi_{i}(B_{min(t_{i+1}, t)} - B_{min(t_{i}, t)})$ and we define Ito integral over $H_{2}^{2}$.
Now I come back to my problem. If I consider $\phi\in H_{2}^{2}$ and I want to compute the expectation of $I(\phi)$ I can find a sequence $(a^n)_{n}\subset\mathcal{R}$ converging to $\phi$ in the $H_{2}^{2}$ norm. Thus I have
$$ \mathbb{E}(I(\phi)_{t}) = \mathbb{E}(\lim_{n\to\infty}I(a^n)_{t}) $$
And now I would like to have the limit outside the expectation. I think it is possible because the operator $I$ from $H_{2}^{2}$ to $L^2$ defined an isometry so $I(a^n)_{t}$ converges in $L^2$ norm to $I(\phi)_{t}$ but this implies the $L_1$ convergence and the expectation being continuous we can put the limit outside the expectation.
Is this correct?
Yes, you have the right idea. We fix $t \geq 0$ and study the mean of the stochastic integral at that time:
\begin{align} |E(I(\phi)_t) - E(I(a_n)_t)| &\leq E|I(\phi)_t - I(a_n)_t| \\ &\leq (E|I(\phi)_t - I(a_n)_t|^2)^{1/2} \\ &\to 0 \end{align} as $n \to \infty$. This last limit holds given that you know that the convergence of $I(a_n)$ to $I(\phi)$ is in $L^2$. Since the mean of $I(a_n)_t$ is zero for all $t$ (more is true: it's a martingale that starts at zero), then it follows that $E(I(\phi)_t)=0$ for all $t \geq 0$.