How to show $0$ is a point of closure of weak topology, but not a limit of weakly covergent sequence in a a subset of $l^2$ 
(von Neumann)For each natural number $n$, let $e_n$ denote the sequence in $\mathcal {l}^2$ whose $n$th component is $1$ and other components vanish. Define$$E = \{e_n + n \cdot e_m : n,m \in \Bbb N \land m > n\}$$

This is an example on page 288, Real Analysis(4ed), Royden et al such that "$0$ is a point of closure with respect to weak topology, of $E$, but there is no sequence in $E$ that converge weakly to $0$". I've no idea how to show this.

The first statement is equivalent to, for any $\epsilon > 0$, and $\{\psi_k\}_{k = 1}^{n} \subset {\mathcal{l}^2}^{\ast}$(${\mathcal{l}^2}^{\ast}$ is the topological dual of ${\mathcal{l}^2}$)
$$\mathcal{N}_{\epsilon, \psi_1, \ldots,\psi_n}(0) = \{x \in \mathcal{l}^2 : |\psi_k(x)|< \epsilon \text{ for} 1 \leq k \leq n\}$$ $$E \cap \mathcal{N}_{\epsilon, \psi_1, \ldots,\psi_n}(0) \neq \varnothing$$
It seems to me it's true not only for $\{\psi_k\}_{k = 1}^{n} \subset {\mathcal{l}^2}^{\ast}$ but also for $\{\psi_k\}_{k = 1}^{n} \subset {\mathcal{l}^2}^{\#}$(the set of all linear functions on $\mathcal{l}^2$). Thus  $\{\psi_k\}_{k = 1}^{n}$ can be represented as $\{(\cdot ,u_k)\}_{k = 1}^{n}$, $u_k \in \mathcal{l}^2 $.
Let $a^k_i$ be the $i$th  component of $u_k$. For any $\epsilon > 0$, we can find $n^k(\epsilon)$ such that for all $i  \geq n^k(\epsilon) $, $a^k_i < \frac{\epsilon}{2}$ and  $a^k_{i^2} < \frac{\epsilon}{2i}$. This may be shown by proof by contradiction, which seems to be messy.
Let $p = \max\{n^k(\epsilon): 1 \leq k \leq n\}$. We have $e_{p}+p \cdot e_{p^2} \in E \cap \mathcal{N}_{\epsilon, \psi_1, \ldots,\psi_n}(0)$.
I got stuck on how to show that there is no sequence in $E$ that converge weakly to $0$.
 A: Weakly convergent sequences are weakly bounded. For a locally convex TVS, weakly bounded coincides with bounded, in this case norm-bounded. Since we're dealing with $l^2$, we have $\lVert e_n + n\cdot e_m\lVert = \sqrt{1+n^2} > n$ for $m > n$.
Thus a weakly convergent sequence $(x_i = e_{n_i} + n_i\cdot e_{m_i})$ in $E$ must have $n_i$ bounded, say $n_i \leqslant k$. Hence there must be an $n_0 \leqslant k$ such that $n_i = n_0$ for infinitely many $i$. Then $\langle x_i \mid e_{n_0} \rangle = 1$ for infinitely many $i$, hence $x_i \not\to 0$ weakly.
To see that $0$ is in the weak closure of $E$, consider $\mathcal{N}_{\varepsilon;\,\psi_1,\,\ldots,\,\psi_n}(0)$. Let $\eta_k = \sum_{i = 1}^n \lvert \psi_i(e_k)\rvert$. Since 1. $\bigl(l^2\bigr)^\ast$ is (anti-)isomorphic to $l^2$, 2. $\eta \in l^2$, and since 3. the coefficients of the $e_j$ are (real and) non-negative for all elements of $E$,
$$\{v \in E \colon \langle \eta\mid v\rangle < \varepsilon\} \subset\mathcal{N}_{\varepsilon;\,\psi_1,\,\ldots,\,\psi_n}(0).$$
So it is sufficient to show that there are $n < m \in \mathbb{N}$ with $\eta_n + n\cdot \eta_m < \varepsilon$. But since $\eta \in l^2$, there is a $k \in \mathbb{N}$ with $\sum_{n = k}^\infty \eta_n^2 < \varepsilon^2/4$, in particular $\eta_k < \varepsilon/2$. Then $\langle\eta\mid e_k + k\cdot e_m\rangle = \eta_k + k\cdot\eta_m \to \eta_k < \varepsilon$, hence $e_k + k\cdot e_m \in \mathcal{N}_{\varepsilon;\,\psi_1,\,\ldots,\,\psi_n}(0)$ for all large enough $m$.
