I'm solving Ex 3.8.1. in Brezis' book of Functional Analysis. Could you have a check on my attempt?
Let $(E, | \cdot |)$ be a Banach space and $d$ a metric on $E$ that induces the weak topology $\sigma(E, E^\star)$. Then there is a sequence $(f_n)$ in $E^\star$ such that each $g\in E^\star$ is a finite linear combination of the $f_n$'s.
My proof: Let $V_n := \{x \in E \mid d(x,0) < n^{-1}\}$ for $n \in \mathbb N^*$. Then there exists a finite sequence $(f_{n, k})_{k=1}^{\varphi(n)}$ in $E^\star$ such that $$V_n \supseteq U_n:=\{x \in E \mid \forall k=1, \ldots,\varphi(n): |\langle f_{n,k}, x \rangle| < 1\}.$$
Fix $f \in E^\star$ and $A := \{x \in E \mid |\langle f, x \rangle| < 1\}$. Let $A_n := A/n := \{x/n \mid x\in A\}$. Then $$\operatorname{ker} f = \bigcap_{n \in \mathbb N^\star} A_n.$$
There exists some $\overline n$ such that $U :=U_{\overline n} \subseteq V_n \subseteq A$. Similarly, we define $U_n :=U/n$. It follows that $$\bigcap_{k=1}^{\varphi(\overline n)} \operatorname{ker} f_{\overline n,k} = \bigcap_{n \in \mathbb N^\star} U_n.$$
It follows from $U \subseteq A$ that $$\bigcap_{k=1}^{\varphi(\overline n)} \operatorname{ker} f_{\overline n,k} \subseteq \operatorname{ker} f.$$
Lemma: Let $f_1, \ldots,f_n,f$ be linear functionals such that $\bigcap_{k=1}^n \ker f_k \subseteq \ker f$. Then $f$ is linearly dependent from $f_1,\ldots,f_n$.
By our lemma, $f$ is a linearly dependent of $\{f_{\overline n,k} \mid k = 1, \ldots, \varphi(\overline n)\}$. It follows that the collection $$\{f_{n, k} \mid n \in \mathbb N^\star, k =1, \ldots, \varphi(n)\}$$ satisfies the requirement.