"Large" compact sets in separable normed space

Question

Let $(X, \lVert \cdot \rVert)$ be a separable normed space. Can we always guarantee that there is a nonempty compact set $K \subseteq B_X$, where $B_X$ is a closed unit ball in $X$ such that:

$$\forall \Lambda \in S_{X^*} \quad \sup_{k \in K} \Lambda(k) > 0,$$

where $S_{X^*}$ denotes the unit sphere in the dual space $X^*$? If not, is it possible to do so, if instead of $S_{X^*}$ we would take some weak* dense subset of $S_{X^*}$?

In the case when $\mathrm{dim}X < \infty$ the answer is yes, since we can just take a closed unit ball for $K$, however, I'm not sure whether this result could be generalized to any separable normed space.

Answer 1

Let $S=\{x_n:n\in\mathbb N\}$ be a dense subset and choose $\varepsilon_n>0$ such that $y_n=\varepsilon_n x_n\to 0$ (e.g., $\varepsilon_n=1/n\|x_n\|$). Then $K=\{ty_n:n\in\mathbb N, |t|=1\}\cup\{0\}$ is compact and for every every $\Lambda\in X^*\setminus \{0\}$ we have $\sup\{\Lambda(k):k\in K\}\ge |\Lambda(y_n)|>0$ for some $n\in\mathbb N$ since otherwise $\Lambda$ would vanish on $S$ and hence on $X$.