Weak* sequential density vs weak* density

Question

Let $X$ be a separable Banach space. Is it true that a weak* dense subspace of $X^{\ast}$ is also weak* sequentially dense?

Answer 1

Let $X=c_0$, and $W\subseteq c_0^*=\ell^1$ be defined by $$W=\{(c_n)_{n\in\mathbb{N}} \in\ell^1: c_0 = \sum_{k=1}^{\infty} c_k\} $$

For $a=(a_n)_{n\in\mathbb{N}} \in c_0$, if $$\sum_{n\in\mathbb{N}} c_na_n=0 \hspace{8mm}\forall c= (c_n)_{n\in\mathbb{N}} \in W$$ then $a=0$. Thus, $W$ is a separating subspace, so weak$^*$ dense in $\ell^1$.

Please see Theorem 1 in the Appendix of the book "Theory of Linear Operations" by Banach, which proves that $W$ is not weak$^*$ sequentially dense (and more).