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?
 A: A basic neighbourhood of $0$ (functional) in the weak-star topology is a finite intersection of sets of the form $O(x,r):=\{f \in X^\ast: |f(x)| < r\}$ with $r>0$ and $x \in X$. This only restricts a functional in such a neighbourhood on finitely many $x \in X$, so if we'd have $\bigcap_{i=1}^n O(x_i,r_i) \subseteq B$ where $B$ is the closed unit ball (in norm!!) in $X^\ast$, it's easy to find some functional $f$, by Hahn-Banach, that is $0$ on all finitely many $x_i$ and (say) $100$ on some $x \notin \{x_1,\ldots,x_n\}$ with $\|x\| = 1$ (we need that $X$ is infinite-dimensional here). Then $\|f\|\ge 100$, so $f \notin B$ while $f \in \bigcap_{i=1}^n O(x_i,r_i)$, contradiction.
So no basic neighbourhood of $0$ (in the weak-star topology) lies inside $B$ and so $0 \notin \operatorname{int}(B)$ as you already suspected to be the case.
A: Your question is answered in the final paragraph of this wikipedia page.

It should be cautioned that despite appearances, the Banach–Alaoglu theorem does not imply that the weak-* topology is locally compact. This is because the closed unit ball is only a neighborhood of the origin in the strong topology, but is usually not a neighbourhood of the origin in the weak-* topology, as it has empty interior in the weak* topology, unless the space is finite-dimensional. 

As pointed out both in your question and by Henno, the crux of the matter is to understand why there is no (weak-*) open neighborhood of the origin contained inside the set $$\mathbb D=\{f\in X^*\colon |f(x)|\leq 1\ \text{ whenever }\ x\in X,\ |x|\leq 1\}.$$
Translating the definition of the weak-* topology to the language of open sets and proceeding as in this answer, one may associate to each (weak-*) open set $U$ containing the origin a finite set $F_U\subset X$ with the property that the set
$$
N=\{f\in X^*\colon |f(x)|<1\ \forall x\in F_U\}
$$
is contained in $U$.
I claim that whenever $\dim(X^*)>|F_U|$, the set $N\setminus \mathbb D$ is non-empty, and in particular when $X^*$ is infinite dimensional it follows from this claim that $\mathbb D$ has empty (weak-*) interior.
The claim holds since $N$ is the intersection of $|F_U|$ slabs, and each time we intersect with a slab we increase the number of bounded dimensions of $N$ by at most one. Thus when the number of slabs is strictly smaller than the dimension of $X^*$, the intersection cannot be bounded.
A: A simple example that I think illustrates this well:
Let $X=H$ be an infinite-dimensional Hilbert space, and $\{e_1, e_2, \dots\}$ an orthonormal set. Let $x_n = 2 e_n$.  If $f \in H^*$ is a continuous linear functional, by the Riesz representation theorem we have $f(x) = \langle x,y \rangle$ for some $y \in H$.  Now by Bessel's inequality we have $\sum_n |\langle x_n, y \rangle|^2 = 4 \sum_n |\langle e_n, y \rangle|^2 < \infty$ and in particular $f(x_n) = \langle x_n, y \rangle \to 0$.  This proves $x_n \to 0$ in the weak-* topology.  If the closed unit ball were a weak-* neighborhood of 0, then it would have to contain all but finitely many of the $x_n$, but it actually contains none of them.
Something like $e_n$ is usually a good first example to consider when thinking about weak or weak-* convergence.
A: 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.
A: 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.
