Let $E$ be a separable or reflexive infinite dimensional normed space, then there exists a sequence in $E'$ weak$^{*}$ convergent to $0$. The following exercise I found in a book in which I found errors, I think that in the following exercise I can remove hypothesis.
Let $E$ be a infinite dimensional normed space, suppose that $E$ is separable or reflexive.  Show that there exists a senquence  $(x_{n}')\subseteq E'$ such that $\left\|x_{n}'\right\|=1$, $n\in\mathbb{N}$, and $x_{n}'\overset{w^{*}}{\rightarrow} 0$  (that is, $x_{n}'(x)\rightarrow 0$ for all $x \in E$).
The problem: I tink that the hypothesis  that $E$ is separable or reflexive can be removed. I want to know if my thinking is correct?
My attepmt:  Assuming we only have $E$ infinite dimensional normed vector space, we denote $S=\left\{x'\in E'\: :\: \left\|x'\right\|=1\right\}$, $\overline{B'}=\left\{x'\in E'\: :\: \left\|x'\right\|\leq 1\right\}$, and $\sigma(E',E)$ the weak$^{*}$ topology in $E'$. We know that 
$$\overline{S}^{\sigma(E',E)}=\overline{B'} \tag{$\bigstar$}$$
where $\overline{S}^{\sigma(E',E)}$ is the closure of $S$ respect to topology $\sigma(E',E)$. (for a proof of ($\bigstar$) see this post ).
Therefore, $0\in \overline{S}^{\sigma(E',E)}$, then there exists $(x_{n}')\subseteq S$  such that $x_{n}'\rightarrow 0$ in the topology $\sigma(E',E)$, that is,    $\left\|x_{n}'\right\|=1$  and $x_{n}'\overset{w^{*}}{\rightarrow} 0$.
Remark: Note that in my attempt I do not use the fact that $E$ is separable or reflexive.
Am I making any mistakes?.

Addendum 
The comment of @GiuseppeNegro is correct, but we know that   $\overline{B}'$ is $\sigma(E', E)$-compact. If $E$ is separable, then we know that  $\overline{B}'$ is metrizable with induced topology of $\sigma(E', E)$. Therefore, my attempt is the proof for the case $E$ separable.
The problem is when $E$ is reflexive. I do not find a relation between reflexivity and metrizable. I'm thinking that when $E$ is reflexive it's not true that $\overline{B}'$ is $\sigma(E', E)$-metrisable, but I can not find a counter-example.
 A: Right, $\overline{B'}$ is $\sigma(E',E)$-compact by Banach-Alaoglu, and if $E$ is separable, then $\sigma(E',E)\rvert_{\overline{B'}}$ is metrisable, so $\overline{B'}$ is also the sequential closure of $S$.
If $E$ is reflexive, we can use the result for separable spaces. Consider a linearly independent sequence $(y_n)$ in $S$, and let $F = \overline{\operatorname{span}} \{ y_n : n \in \mathbb{N}\}$. Then $F$ is a separable closed subspace of $E'$. As a closed subspace of a reflexive normed space, $F$ is itself reflexive, so $F = (F')'$ (with the usual abuse of notation). The separability of $(F')'$ implies the separability of $F'$, so by the first part we know that there is a sequence $(x_n)$ in $S \cap F$ such that $x_n \to 0$ in $\sigma(F,F')$.
It remains to see that $\sigma(F,F') = \sigma(E',E)\lvert_F$. The continuity of the inclusion $F \hookrightarrow E'$ gives $\sigma(E',E'')\lvert_F \subset \sigma(F,F')$. For the other inclusion, consider $\lambda \in F'$. By Hahn-Banach, there is a $\mu \in E''$ with $\lambda = \mu\lvert_F$. Now
$$\{ f \in F : \lvert \lambda(f)\rvert < \varepsilon\} = \{ x \in E' : \lvert \mu(x)\rvert < \varepsilon\} \cap F$$
is immediate, so every $\sigma(F,F')$-neighbourhood of $0$ is also a $\sigma(E',E'')\lvert_F$-neighbourhood of $0$. By reflexivity of $E$, $\sigma(E',E'') = \sigma(E',E)$. Thus $x_n \to 0$ in $\sigma(E',E)$.
I haven't yet come up with an example of a non-separable non-reflexive Banach space in which there is no sequence $(x_n)$ in $S$ with $x_n \to 0$ in $\sigma(E',E)$. Will think about that further after sleeping.
I still haven't found an example. For such an example, $E$ must not have an infinite-dimensional separable (or reflexive) complemented subspace, which rules out spaces like $c_0(T)$ or $\ell^1(T)$ for uncountable $T$. If $E \cong F \oplus G$ with separable (or reflexive) infinite-dimensional $F$, then the identification $F' \cong G^{\perp} = \{\lambda \in E' : G\subset \ker \lambda\}$ gives a sequence in $S \cap G^{\perp}$ that converges weakly to $0$. I suspect that $\ell^{\infty}(\mathbb{N})$ would work, but so far I cannot prove (or disprove) that.
