On the weak closure

Let $\lbrace e_n \rbrace$ for the standard unit vectors in $l_2$.

I want to show that $0$ is in weak closure of $\lbrace\sqrt{n}e_n\rbrace$ but no subsequence of $\lbrace \sqrt{n}e_n\rbrace$ weakly converges to 0.

For the second assertion, I have the following answer.

If $\lbrace \sqrt{n_k}e_{n_k}\rbrace$ be weakly convergent subsequence, then it must be norm-bounded. However, $\| \sqrt{n_k}e_{n_k}\|=\sqrt{n_k}\to \infty$ as $k\to \infty$ which is a contradiction.

However, for the first assertion, I cannot figure it out that the difference between the condition that 0 lies in weak closure of sequence and the condition that there is a subsequence weakly converges to 0.

Thanks.

The hint from Pedersen, Analysis Now: If $x = \sum \alpha_n e_n$ in $l_2$, then there is no $\epsilon > 0$ such that $|(y|x)| > \epsilon$ for every $y = \sqrt{n} e_n$.

Problem 28 of Halmos's A Hilbert space problem book says:

The weak topology of an infinite-dimensional Hilbert space is not metrizable.

It comes with a warning, "The shortest proof of this is tricky." Then there is the hint, "Construct a sequence that has a weak cluster point but whose norms tend to $\infty$." Finally, there is the solution, which begins by stating and solving your problem. The example is attributed by Halmos to Shields.