If $X$ is separable then there exists $D$ a countable symmetric subset of the unit ball $B^*$ in $X^*$ dense with respect to the Mackey topology Let $X$ be a separable Banach space. $X^*$ will denote the dual of $X$. $B^*$ will denote the unit ball in $X^*$.
$X^*$ equipped with the Mackey topology $\tau(X^*,X)$, i.e. the strongest, locally convex topology $\tau$ on $X^∗$ for which we have $(X^*,\tau )^∗ = X$.
Why there exists a countable subset $D$ of the unit ball $B^*$, symmetric $(D=-D)$ such that:
$$
B^*=\overline{D}^{\tau(X^*,X)}
$$
 A: There are two key ideas in here: one is a basic sequence of thoughts from functional analysis:  $X$ is separable so $X^*$ is metrisable, and by the Banach-Alaoglu theorem the unit ball $B_{X^*}$ is compact and so is a compact metric space.  Compact metric spaces are separable as well, so we can find a (countable) dense subset of $B_{X^*}$.  This tells us that looking for your subset $D$ has hopes of success.
The second idea is that since $X$ is separable its unit ball $B_X$ contains a dense sequence and we can use this to construct something that has the required properties.  So, let $S=\{x_n\}$ be such a dense sequence.  Since we're talking about the Mackey topology we know that there is a neighbourhood base that is the family of polars of weakly-compact balanced (symmetric) convex subsets of $X$, so let us set 
$$ V^\circ_{X^*} = \{x^* \in X^* \mid |x^*(x_n)|\leq 1 \ \forall x_n \in S\} $$
to be the polar of $S$. It is clear that this is a subset of $B_{X^*}$ and we claim that in fact they are equal:
Suppose $y^* \in V^\circ_{X^*} \setminus B_{X^*} $.  Then $\|y^*\|\gt 1$ and $|y^*(x_n)| \leq 1 \ \forall x_n \in S$  This means that for $x\in B_X$ we have $|y^*(x)| - |y^*(x_n)| \geq |y^*(x_n)-y^*(x)| = |y^*(x_n-x)| \gt \varepsilon \gt 0$.  But we also know that $\|y^*(x_n-x)\| \leq \|y^*\|\cdot\|x_n-x \| < \varepsilon$, which is impossible.  So no such $y^*$ exists and $V^\circ_{X^*} = B_{X^*}$.
So now we choose our symmetric countable dense subset $D$ from $V^\circ_{X^*}$ where the symmetry property is available to us because the polar is symmetric.
