About sequential weak closure in $\ell^2$ I'm having some troubles with the following exercise:

Show that the set $$A= \left(\sqrt{n}e_n \mid n\in \mathbb N \right)$$ is sequentially weakly closed in $\ell^2$.

My idea is to take an arbitrary weakly convergent sequence $\left(a_n\right) \subset A$ and then to show that the weak limit is inside $A$. I do know that a weakly convergent sequence $\left(a_n\right)$ converges weakly to $a$  if and only if the following holds: $ \lim_{n \to +\infty} \langle x^*, a_n\rangle = \langle x^*, a\rangle\, (\forall x^* \in X^*)$ , where $X^*$ denotes the dual of $\ell^2$ which is isomorphic to $\ell^2$ itself by Riesz.
Can someone give me any hints?
 A: I'd suggest you prove that any weakly convergent sequence $(\sqrt{n_i} e_{n_i})_{i \in \mathbf{N}}$ in $A$ is eventually constant. By the Uniform Boundedness Principle, any sequence weakly convergent sequence is bounded, say from above by $n_0 \in \mathbf{N}$. Let $x = \sum_{j = 1}^{n_0} \sqrt{j} \cdot e_j$, then $\langle x, \sqrt{n_i} e_{n_i} \rangle = n_i$ converges as $i \rightarrow \infty$. This is only the case, when $n_i$ is eventually constant (pick $\varepsilon = \frac{1}{2}$). 
A: Ok I think I might have found a solution thanks to the reminder by Steven that any weak convergent sequence (which I will proof below for the sake of completness) is bounded:
The idea is to pick any sequence $(a_i)_{i\in \mathbb N}:= (\sqrt{n_i}e_{n_i})_{i \in \mathbb N}$ in $A$ such that it has a weak limit, say a, in $\ell^2$. 
For notational reasons: $X:=\ell^2$;
Observe the following:
$$lim_{i\to \infty}\iota(a_i)(x^*)=lim_{i \to \infty}\langle x^*,a_i\rangle =\langle x^*,a \rangle= \iota(a)(x^*) \space (\forall x^* \in X^*)$$
where $\iota: X \to X^{**}$ is the standard isometric embedding. Since $\iota(a_i)(x^*)$ converges for every linear functional $x^* \in X^*$ it follows that it's bounded (sequence converges $\implies$ sequence bounded) for every $x^* \in X^*$, hence by Banach-Steinhaus (Uniform Boundedness Principle) it follows that
$$sup_{i\in \mathbb N}\Vert \iota(a_i) \Vert < \infty$$
and since $\iota$ is isometric it follows that:
$$sup_{i \in \mathbb N}\Vert \iota(a_i)\Vert=sup_{i \in \mathbb N}\Vert a_i\Vert=sup_{i \in \mathbb N}\Vert \sqrt{n_i}e_{n_i}\Vert=sup_{i \in \mathbb N}\vert n_i \vert< \infty$$ 
Thus the sequence $(\sqrt{n_i}e_{n_i})_{i \in \mathbb N}$ is eventually constant, hence the weak limit $a$ is in $A$. q.e.d.
