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?

The open unit ball in the strong topology is NOT open in the weak$^*$ topology. The weak$^*$ topology is way coarser than the strong one (less open sets).
To get an idea, consider $\ell^\infty (\mathbb N)$. A local basis of open neighbourhoods in the strong topology are the sets
$$
U_n=\left\{(a_k)_{k\in\mathbb N}: |a_k|< 1/n,\,\,\text{for all $k\in\mathbb N$} \right\}, \,\,n\in\mathbb N.
$$ 
In the case of the weak$^*$ topology, a local basis of open neighbourhoods are finite intersections of the sets
$$
W_{m,n}=\left\{(a_k)_{k\in\mathbb N}: |a_m|\le 1/n\right\}, \,\,m,n\in\mathbb N.
$$ 
Each $\overline U_n=\left\{(a_k)_{k\in\mathbb N}: |a_k|\le 1/n,\,\,\text{for all $k\in\mathbb N$} \right\}$ is compact in the weak$^*$ sense, but its interior is NOT open in the strong sense.
