I have the next unitary representation, $\pi : G\rightarrow \mathcal{U}(H)$, where G is a closed subgroup of $S_{\infty}$ (the group of bijective functions from $\mathbb{N}\rightarrow \mathbb{N}$), and $\mathcal{U}(H)$ is the group of unitary operators on a separable hilbert space $H$, and I want to show that $\bigcup\{H_V : V \text{ clopen subgroup of } G\}$ is dense in $H$, where $H_V=\{ x \in H : \pi(v)x=x, \forall v \in V \}$.
I think it's easy but I am not sure, I take a sequence $x_n \in H_V$, and assume that $x_n \rightarrow x$, where $x \in H$, then because $\pi$ is a unitary operator, it's bounded and thus continuous as well, so we know that $\pi(v)x_n \rightarrow \pi(v)x$, and because $\pi(v)x_n=x_n$, we get from uniqueness of the limit that $\pi(v)x=x$, so $x\in H_V$.
Where did I get it wrong?