Suppose $(x_n)$ is bounded and for all $e \in E$, $ \to 0$ as $n \to \infty$.Does this implies that $ \to 0$ for all $y$ in $H$? I'm reading a solution in which following result is used.

Suppose $H$ be a hilbert space and $E$ be a ONB for H.Suppose $(x_n)$ is bounded and $<x_n,e>  \to 0$ as $n \to \infty$ for every $e \in E$.Does this implies that $<x_n,y> \to 0$ for all $y$ in $H$ ?

I think it is true but unable to prove .Please help.
 A: Let $||x_n||<A$
Take $M$ to be the subspace of finite linear combination of basis elements. Then it is easy to see that $M$ is dense in $H.$ $\langle x_n,e\rangle \rightarrow 0 \forall e\implies \langle x_n,m\rangle \rightarrow 0 \forall m\in M$
Now fix $y\in H$. Take $\epsilon >0$
Choose $m\in M$ such that $||m-y||<\epsilon A^{-1} $ 
As $\langle  x_n,m\rangle \rightarrow 0  $  thereesists $k>0$ such that $$|\langle x_n,e\rangle |<\epsilon$$ for all $n>k$ 
Now 
$$
|\langle x_n,y \rangle|\leq |\langle x_n,y-m \rangle|+|\langle x_n,m \rangle|\leq ||x_n||||y-m||+\langle x_n,m \rangle|<A \epsilon \frac{1}{A}+\epsilon=2\epsilon $$
for all $n>k$
A: Let $V := \langle E\rangle $ be the span of $E $, i.e. the set of finite linear combinations of the orthonormal basis. It is easy to see $\langle x_n, v\rangle \to 0$. 
Now, let $y \in H $ be arbitrary and $\epsilon >0$. There is $v \in V $ with $\Vert v -y\Vert <\epsilon $, since $V $ is dense because $E $ is an orthonormal basis. Now,
$$
|\langle x_n, y\rangle | \leq |\langle x_n, y-v\rangle| + |\langle x_n, v\rangle | \leq C \|y-v\| + |\langle x_n ,v\rangle| \leq C\epsilon +\epsilon
$$
for $n $ large.
Thus, $\langle x_n, y\rangle \to 0$.
Note that the last part of the proof is the usual argument for showing that weak convergence of a bounded sequence on a dense subset implies weak convergence on the whole space.
