I was curious whether the following property for separable spaces characterizes them.
Let $X$ be a separable Banach space, and let $B^*$ be the closed unit ball of $X^*$. Then there exists a countable set $D \subset B^*$ such that for all $x\in X$:
$$ ||x|| = \sup_{x^*\in D} |\langle x,x^*\rangle|$$
If $\{x_i\} \subset X$ is a countable dense subset of $X$ and $x_i^*\in B^*$ are such that $\langle x,x^*\rangle=||x_i||$ (note that $x_i^*$ exists by Hahn-Banach), then $D=\{x_i^*\}$ is such a set.
My question is if such a property characterizes separable spaces. This is, if there exists a countable set $D\subset B^*$ such that for all $x\in X$:
$$ ||x|| = \sup_{x^*\in D} |\langle x,x^*\rangle|$$ then must $X$ be separable?
The answer is NO. Let $X=\ell^{\infty}$ and $x^{*}_n(x)=x_n$. Then $x_n^{*}$ is a continuous linear functioinal on $X$ and $\|x\|=\sup | \langle x, x_n^{*} \rangle|$ for all $x \in X$ but $X$ is nor separable.