Dual space $E'$ is metrizable iff $E$ has a countable basis

I saw that it was already asked, but the book where I’m studying is slightly different. Recall some definition, if $E$ it’s $\mathbb{K}$-vector space and let $\mathcal{E}$ be a vector subspace of the algebraic dual of $E$, wich is the vector space of all linear forms on $E$. We say that $(E,\mathcal{E})$ is a dual couple if $\mathcal{E}$ separates points of $E$, that is, $u(x)=u(y)$ $\forall u \in \mathcal{E}$ implies $x=y$. We have the following result:

  1. “Suppose $(E,\mathcal{E})$ is a dual couple , $u_1,…,u_n , u \in \mathcal{E}$.
    Then $u=\sum_{k=1}^n \lambda_k u_k$ iff $u_1(x)=…=u_n(x)=0$ implies $u(x)=0$”

Exercise (a): Let $E$ be an infinite-dimensional locally convex space. Prove that the weak* topology $\sigma(E’,E)$ on $E’$ is metrizable if and only if $E$ has a countable algebraic basis.

Hint (a): Let $U_n=\lbrace \max\lbrace p_{x_1},…,p_{x_N} \rbrace < 1/M \rbrace$. If $x \in E$, every ball $\lbrace p_x < \epsilon \rbrace$ contains some $U_n$; that is, $|u(x_j| < 1/n$ $\forall j$ implies $|u(x)| <1$. Hence $u(x_1)= \cdot \cdot \cdot u(x_N)=0$ implies $u(x)=0$ and $x \in \mathrm{span}(x_1,…,x_n)$.

Proof (Exercise (a)).
We assume that the weak* topology $\sigma(E’,E)$ on $E’$ it’s metrizable. Let $x_1,…,x_N \in E \setminus \lbrace 0 \rbrace$. Note that the weak* topology is defined by seminorm family $\mathcal{F}= \lbrace p_x(u)=|u(x)| : x \in E \rbrace$, where
\begin{align*}
\displaystyle U_n := \lbrace u \in E’ : \max \lbrace p_{x_1}(u),…,p_{x_N}(u) \rbrace < 1/n \rbrace
\end{align*}
It is an open neighborhood in weak* topology.
By hypothesis $\forall x \in E$, $\exists n \in \mathbb{N}$ and exists open ball $B_{E’}:=\lbrace u \in E’ : p_x(u):=|u(x)| < \epsilon \rbrace$ such that $U_n \subset B_{E’}$.
Now, if $u \in U_n$ we have $|u(x_j)| < 1/n$ $\forall j$ and then $|u(x)|< 1$ with $\epsilon=1$, and by (1) we have:
\begin{align*}
u(x_1)=…=u(x_N)=0 \Longrightarrow 0=u(x)=\sum_{j=1}^n \lambda_j u(x_j)
\end{align*}
by linearity $x \in \mathrm{span}(x_1,…,x_n)$.
Reciprocally, just reverse the above implications, using results in (1)?

Someone can help me? can you check if proof of exercise (a) is correct?
Thank you.

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Filling in some details into your argument above yields the following:

Assume that the weak$^*$-topology on $E’$ is metrizable, then there is a countable basis $\{V_n\}$ of neighbournoods of zero. Each of these contains a set of the form
$$U_n:=\{u\in E’\colon \max\{p_{x_{n,1}}, \ldots, p_{x_{n,\varepsilon_n}}\} < 1/M_n\},$$
where $N_n,M_n\in\mathbb{N}$, since the latter is a basis of neighbourhoods of zero for the weak$^*$-topology.
So for every $x\in E$ and every $\varepsilon>0$, the set
$$
\{u\in E’\colon |u(x)|<\varepsilon\}
$$
contains a set $U_n$. Hence $u(x_{n,1})=\ldots u(x_{n,N_n})=0$ implies $u(x)=0$ and therefore $x\in\operatorname{span}\{x_{n,1},\ldots,x_{n,N_n}\}$. Therefore
$$
\bigcup_{n\in\mathbb{N}}\{x_{n,1},\ldots,x_{n,N_n}\}
$$
is a countable basis of $E$.

To prove the other implication, let $\{x_n\}$ be a countable basis of $E$ and set
$$
U_{m,n} := \left\{u\in E’\colon \max_{1\leq i\leq n} |u(x_i)| < \frac{1}{m}\right\}.
$$
We have to show that $\{U_{m,n}\colon m,n\in\mathbb{N}\}$ is a basis of neighbourhoods of zero. Let $x\in E$, then there is an $N\in\mathbb{N}$ such that $x=\sum_{i=1}^{N} \lambda_i x_i$ for some $\lambda_i$. We get
$$
|u(x)| \leq \sum_{i=1}^{N} |\lambda_i| |u(x_i)| \leq C \max_{1\leq i \leq N} |u(x_i)|,
$$
where $C:=\sum_{i=1}^{N} |\lambda_i|$. Hence every set of the form
$$
\{u\in E’\colon |u(x)| < \varepsilon\}
$$
and therefore also finite intersections of these, contains a set of the form $U_{n,m}$.