For a separable Banach space $X$, the unit sphere of $X^*$ always contains a countable set $D$ such that $$ \left\Vert x \right\Vert = \sup_{f \in D} \left\vert f(x) \right\vert \qquad \mbox{ for every }x\in X.$$ Assume that $A\subset X$ is nonempty, bounded, convex, and $\inf_{a\in A}\left\Vert a \right\Vert > 1$. Then, by Hahn-Banach separation theorem, we can find $f\in X^*$, $\left\Vert f \right\Vert = 1$, such that $f(a) > 1$ for all $a\in A$.
Is it possible to find also $f\in D$ such that $f(a)>1$ (or $f(a)>1/2$) for all $a\in A$?
Not in general. Here is a counterexample.
Let $X = C([-1,1])$. For $t \in [-1,1]$, let $\delta_t$ denote the point mass / evaluation functional $\delta_t(x) = x(t)$. Let $D = \{ \delta_q : q \in [-1,1] \cap \mathbb{Q}\}$. Then $D$ is countable and we have $\|x\| = \sup_{f \in D} |f(x)|$ for every $x \in X$.
Let $$y(t) = \begin{cases} 4t, & -1 \le t \le 0 \\ 0, & 0 \le t \le 1 \end{cases}$$ and $$z(t) = -y(-t) = \begin{cases} 0, & -1 \le t \le 0 \\ 4t, & 0 \le t \le 1 \end{cases}$$ Let $A = \{s y + (1-s)z : 0 \le s \le 1\}$ be the line segment connecting $y$ and $z$. $A$ is nonempty, convex and bounded (even compact). And for any $a \in A$ we have either $a(-1) \le -2$ or $a(1) \ge 2$, so $\inf_{a \in A} \|a\| = 2 > 1$. But for any $f = \delta_q \in D$, if $-1 \le q \le 0$ we have $f(z)=0$ and if $0 \le q \le 1$ we have $f(y) = 0$. So in all cases there exists $a \in A$ with $f(a) = 0$.