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A partial answer. Suppose the weak$^*$ sequential closure of $S^*$ contains $0$. So there is a sequence $(f_n) \subseteq X^*$ with $\|f_n\|=1$ for each $n$, and with $f_n\rightarrow 0$ weak$^*$. For $f\in B^*$ the unit ball, we seek a bounded sequence $(t_n)$ of scalars such that $f+t_nf_n \in S^*$ for each $n$, as then $f+t_nf_n \rightarrow f$ weak$^*$. We can find such $t_n$ from the triangle inequality, $$ \big| \|f\| - |t_n| \big| \leq \|f + t_nf_n\| \leq \|f\| + |t_n|, $$ and using that $t\mapsto \|f+tf_n\|$ is continuous. Conclude: the weak$^*$-sequential closure of $S^*$ is all of $B^*$.

So, when is $0$ in the weak$^*$ sequential closure of $S^*$?

Here is an easy argument if $X$ is separable. Then $X$ hash a countably dense subset $\{x_k\}$, then for each $n$ by Hahn-Banach we can find $f_n\in S^*$ with $f_n(x_k)=0$ for $k\leq n$. Given $x\in X$ and $\epsilon>0$ there is $k$ with $\|x-x_k\|<\epsilon$, and so for $n\geq k$ we have $|f_n(x)| = |f_n(x-x_k)| < \epsilon$. Conclude that $f_n\rightarrow 0$ weak$^*$.

As Dirk Werner points out, in the general case, we can use the Josefson-Nissenzweig Theorem which exactly says that any (infinite-dimensional) Banach space has the property we need: $X^*$ has a weak$^*$-null sequence of norm one vectors. Here is one source for this theorem: Extract from Diestel, Sequences and Series in Banach Spaces

Suppose the weak$^*$ sequential closure of $S^*$ contains $0$. So there is a sequence $(f_n) \subseteq X^*$ with $\|f_n\|=1$ for each $n$, and with $f_n\rightarrow 0$ weak$^*$. For $f\in B^*$ the unit ball, we seek a bounded sequence $(t_n)$ of scalars such that $f+t_nf_n \in S^*$ for each $n$, as then $f+t_nf_n \rightarrow f$ weak$^*$. We can find such $t_n$ from the triangle inequality, $$ \big| \|f\| - |t_n| \big| \leq \|f + t_nf_n\| \leq \|f\| + |t_n|, $$ and using that $t\mapsto \|f+tf_n\|$ is continuous. Conclude: the weak$^*$-sequential closure of $S^*$ is all of $B^*$.

So, when is $0$ in the weak$^*$ sequential closure of $S^*$?

Here is an easy argument if $X$ is separable. Then $X$ hash a countably dense subset $\{x_k\}$, then for each $n$ by Hahn-Banach we can find $f_n\in S^*$ with $f_n(x_k)=0$ for $k\leq n$. Given $x\in X$ and $\epsilon>0$ there is $k$ with $\|x-x_k\|<\epsilon$, and so for $n\geq k$ we have $|f_n(x)| = |f_n(x-x_k)| < \epsilon$. Conclude that $f_n\rightarrow 0$ weak$^*$.

As Dirk Werner points out, in the general case, we can use the Josefson-Nissenzweig Theorem which exactly says that any (infinite-dimensional) Banach space has the property we need: $X^*$ has a weak$^*$-null sequence of norm one vectors. Here is one source for this theorem: Extract from Diestel, Sequences and Series in Banach Spaces

A partial answer. Suppose the weak$^*$ sequential closure of $S^*$ contains $0$. So there is a sequence $(f_n) \subseteq X^*$ with $\|f_n\|=1$ for each $n$, and with $f_n\rightarrow 0$ weak$^*$. For $f\in B^*$ the unit ball, we seek a bounded sequence $(t_n)$ of scalars such that $f+t_nf_n \in S^*$ for each $n$, as then $f+t_nf_n \rightarrow f$ weak$^*$. We can find such $t_n$ from the triangle inequality, $$ \big| \|f\| - |t_n| \big| \leq \|f + t_nf_n\| \leq \|f\| + |t_n|, $$ and using that $t\mapsto \|f+tf_n\|$ is continuous. Conclude: the weak$^*$-sequential closure of $S^*$ is all of $B^*$.

So, when is $0$ in the weak$^*$ sequential closure of $S^*$? If $X$ is separable, say with a countably dense subset $\{x_k\}$, then for each $n$ by Hahn-Banach we can find $f_n\in S^*$ with $f_n(x_k)=0$ for $k\leq n$. Given $x\in X$ and $\epsilon>0$ there is $k$ with $\|x-x_k\|<\epsilon$, and so for $n\geq k$ we have $|f_n(x)| = |f_n(x-x_k)| < \epsilon$. Conclude that $f_n\rightarrow 0$ weak$^*$.

I am afraid I do not know what happens in the non-separable case. The non-separable spaces $\ell^p(\Gamma)$ have the same property.

A partial answer. Suppose the weak$^*$ sequential closure of $S^*$ contains $0$. So there is a sequence $(f_n) \subseteq X^*$ with $\|f_n\|=1$ for each $n$, and with $f_n\rightarrow 0$ weak$^*$. For $f\in B^*$ the unit ball, we seek a bounded sequence $(t_n)$ of scalars such that $f+t_nf_n \in S^*$ for each $n$, as then $f+t_nf_n \rightarrow f$ weak$^*$. We can find such $t_n$ from the triangle inequality, $$ \big| \|f\| - |t_n| \big| \leq \|f + t_nf_n\| \leq \|f\| + |t_n|, $$ and using that $t\mapsto \|f+tf_n\|$ is continuous. Conclude: the weak$^*$-sequential closure of $S^*$ is all of $B^*$.

So, when is $0$ in the weak$^*$ sequential closure of $S^*$?

Here is an easy argument if $X$ is separable. Then $X$ hash a countably dense subset $\{x_k\}$, then for each $n$ by Hahn-Banach we can find $f_n\in S^*$ with $f_n(x_k)=0$ for $k\leq n$. Given $x\in X$ and $\epsilon>0$ there is $k$ with $\|x-x_k\|<\epsilon$, and so for $n\geq k$ we have $|f_n(x)| = |f_n(x-x_k)| < \epsilon$. Conclude that $f_n\rightarrow 0$ weak$^*$.

As Dirk Werner points out, in the general case, we can use the Josefson-Nissenzweig Theorem which exactly says that any (infinite-dimensional) Banach space has the property we need: $X^*$ has a weak$^*$-null sequence of norm one vectors. Here is one source for this theorem: Extract from Diestel, Sequences and Series in Banach Spaces

A partial answer. Suppose the weak$^*$ sequential closure of $S^*$ contains $0$. So there is a sequence $(f_n) \subseteq X^*$ with $\|f_n\|=1$ for each $n$, and with $f_n\rightarrow 0$ weak$^*$. For $f\in B^*$ the unit ball, we seek a bounded sequence $(t_n)$ of scalars such that $f+t_nf_n \in S^*$ for each $n$, as then $f+t_nf_n \rightarrow f$ weak$^*$. We can find such $t_n$ from the triangle inequality, $$ \big| \|f\| - |t_n| \big| \leq \|f + t_nf_n\| \leq \|f\| + |t_n|, $$ and using that $t\mapsto \|f+tf_n\|$ is continuous. Conclude: the weak$^*$-closure of $S^*$ is all of $B^*$.

So, when is $0$ in the weak$^*$ sequential closure of $S^*$? If $X$ is separable, say with a countably dense subset $\{x_k\}$, then for each $n$ by Hahn-Banach we can find $f_n\in S^*$ with $f_n(x_k)=0$ for $k\leq n$. Given $x\in X$ and $\epsilon>0$ there is $k$ with $\|x-x_k\|<\epsilon$, and so for $n\geq k$ we have $|f_n(x)| = |f_n(x-x_k)| < \epsilon$. Conclude that $f_n\rightarrow 0$ weak$^*$.

I am afraid I do not know what happens in the non-separable case. The non-separable spaces $\ell^p(\Gamma)$ have the same property.

A partial answer. Suppose the weak$^*$ sequential closure of $S^*$ contains $0$. So there is a sequence $(f_n) \subseteq X^*$ with $\|f_n\|=1$ for each $n$, and with $f_n\rightarrow 0$ weak$^*$. For $f\in B^*$ the unit ball, we seek a bounded sequence $(t_n)$ of scalars such that $f+t_nf_n \in S^*$ for each $n$, as then $f+t_nf_n \rightarrow f$ weak$^*$. We can find such $t_n$ from the triangle inequality, $$ \big| \|f\| - |t_n| \big| \leq \|f + t_nf_n\| \leq \|f\| + |t_n|, $$ and using that $t\mapsto \|f+tf_n\|$ is continuous. Conclude: the weak$^*$-sequential closure of $S^*$ is all of $B^*$.

So, when is $0$ in the weak$^*$ sequential closure of $S^*$? If $X$ is separable, say with a countably dense subset $\{x_k\}$, then for each $n$ by Hahn-Banach we can find $f_n\in S^*$ with $f_n(x_k)=0$ for $k\leq n$. Given $x\in X$ and $\epsilon>0$ there is $k$ with $\|x-x_k\|<\epsilon$, and so for $n\geq k$ we have $|f_n(x)| = |f_n(x-x_k)| < \epsilon$. Conclude that $f_n\rightarrow 0$ weak$^*$.

I am afraid I do not know what happens in the non-separable case. The non-separable spaces $\ell^p(\Gamma)$ have the same property.