Preliminaries: Let $X$ be a separable Banach space, $X^*$ it's dual of linear, bounded functionals and $\overline{B}_1^* \subset X^*$ the dual unit ball which is closed w.r.t. the dual norm. A sequence $(x^*_n) \subset X^*$ converges to $x^*\in X^*$ w.r.t. to the $w^*$-topology iff $x_n^*(x) \to x^*(x)$ f.a. $x \in X$.
Question: I want to prove that there is a sequence $(x^*_n) \subset X^*$ which closure w.r.t. to the $w^*$-topology is the closed dual ball, i.e. $$\overline{\{x^*_n\}}^{w^*} = \overline{B}_1^*.$$ I think I got the $"\subseteq"$-direction but need help with the other one. Also, I wrote down anything I know, so maybe someone can point out some errors to me.
My attempt: We know that by Alaoglu $\overline{B}_1^*$ is $w^*$-compact. The separability of $X$ implies that there is a metric such that the induced topology coincides with $w^*$. So $\overline{B}_1^*$ is $w^*$-sequentially compact, which means that every sequence has a $w^*$-convergent subsequence. Let $(x_n)\subset X$ be a dense sequence of $X$. By Hahn-Banach there is a norming unit sequence $(x^*_n) \subset X^*$, i.e. $\|x^*_n\|_*=1$ and $x^*_n(x_n)=\|x_n\|$, which will be our candidate for the above mentioned relation.
$"\subseteq":$ Let $x^*\in \overline{\{x^*_n\}}^{w^*}$ which is a limit point of a subsequence $(x^*_{n_k})$ w.r.t. $w^*$-topology, which means $x^*_{n_k}(x) \to x^*(x)$ f.a. $x\in X$. Then we get by another corollary of Hahn-Banach and the boundness that $$\|x^*\|_*= \sup_{\|x\|=1} |x^*(x)| = \sup_{\|x\|=1} \lim_k |x^*_{n_k}(x)| \leq \lim_k \sup_{\|x\|=1} |x^*_{n_k}(x)| \leq \lim_k \sup_{\|x\|=1} \|x^*_{n_k}\|_*\|x\| =1.$$ So therefore $x^* \in \overline{B}_1^*$.
As I said I don't know how to chose for an arbitrary $x^* \in X^*$ a $w^*$-convergent supsequence of $\overline{\{x^*_n\}}^{w^*}$. I know that we can approximate every $x \in X$ and therefore find a subsequence of $\{x^*_n\}$ with $x^*_{n_{k_l}}(x) \to \|x\|$, but i can't see how this would help. Thanks in advance
