Why is $X^*$ with the weak-* topology not a locally compact vector space? I know that locally compact Hausdorff topological vector space must be isomorphic to $\mathbb{C}^d$ or $\mathbb{R}^d$ for some $d\in \mathbb{N}$.
However the Banach Alaouglu theorem says that the closed unit ball of $X^*$ is compact with respect to the weak-* topology, which seemed to me at first sight to mean that  it is locally compact. But I know that this space with the weak-* topology is also Hausdorff, and allegedly should mean that $X^*$ with the weak-* topology is finite dimensional.
My question is then why is the closed unit ball in $X^*$ not a compact neighbourhood of $0$?
I found this thread saying that exactly, but I do not understand why. Is the unit ball of $X^*$ not a neighbourhood of $0\in X^*$ with the weak-* topology, which is contained in the aforementioned set?

Every basic neighbourhood of zero in the $w^*$ topology of $X^*$ is of the form
$$U = \{x^* \in X^*; |x^*(x_1)| < \varepsilon, \dots, |x^*(x_n)| < \varepsilon\}$$ 
for some $x_1, \dots , x_n \in X$ and $\varepsilon > 0$.
Hence, if $X$ is infinite dimensional, every such $U$ is unbounded as it contains a subspace of $X^*$ of codimension at most $n$.
Then $B_{X^*}$ has empty $w^*$ interior as otherwise it would have to contain a basic set but bounded sets can't contain unbounded sets.
