The weak$^*$ topology on $X^*$ is not first countable if $X$ has uncountable dimension. I learnt without proof that if $X$ is a normed space of uncountable dimension, then the weak* topology on $X^*$ is not first countable. Can anyone point out how I should go about proving it?
I tried the following:
Suppose that the weak* topology on $X^*$ is first countable. Then we may take a sequence $(x_n)$ such that 
$$
V_n=\big\{x^*\in X^*: \lvert x^*(x_i)\rvert<1, \,\,\text{for all}\,\, i=1, 2, \dots, n\big\}
$$ 
gives a (descending sequence of) countable base at $0\in X^*$. Because I want to find a contradiction, I guess I should try to show that $(x_n)$ span $X$.
Is the above attempt getting anywhere?
 A: Let $V_n$, $x_n$ be as in the posted question, and let $\iota : X \to X^{**}$ be the natural embedding. The idea of the proof comes from the discussion in the comment field above. 
Take $x\in X$. $$A:=\{x^*\in X^*: |x^*(x)|<1\}$$ is a weak* neighbourhood of $0$ in $X^*$. Therefore, for some $n$, $\bigcap_{1\le i\le n}\ker \iota(x_i)\subset V_n\subset A$. So if $x^*\in \bigcap_{1\le i\le n}\ker \iota(x_i)$, then for all scalar $\lambda$, $\lambda x^*\in \bigcap_{1\le i\le n}\ker \iota(x_i) \subset A$. So $|(\lambda x^*)(x)|<1$. This forces $x^*(x)=0$. So $\bigcap_{1\le i\le n}\ker \iota(x_i)\subset \ker \iota(x)$. So $x\in\langle x_1, \dots, x_n\rangle$. So $(x_n)_{n\in\mathbb{N}}$ is a Hamel basis for $X$, which is a contradiction. 
A: In the weak$^*$ topology a sub-base of the neighborhoods of $0$ is obtained by
sets of the form
$$
W_{x,\varepsilon}=\big\{x^*\!\in X^*: \lvert x^*(x)\rvert<\varepsilon\big\}, 
\quad \varepsilon>0,\, x\in X,
$$
and a local base of 0  (in the weak$^*$ topology) is obtained by finite intersections
of the above sets.
In particular, if ${\mathcal N}$ is a base of the neighbourhoods of $0\in X^*$,
then for every $U\in\mathcal N$, there exist $n\in\mathbb N$, $x_1, \ldots x_n\in X$
and $k_1,\ldots k_n\in\mathbb N$, such that
\begin{equation}
W_{x_1,1/k_1}\cap\cdots\cap W_{x_n,1/k_n}\subset U. \tag{1}
\end{equation}
In fact, if each $U$ in $\mathcal N$ is replaced by an intersection
$W_{x_1,1/k_1}\cap\cdots\cap W_{x_n,1/k_n}$ satisfying (1),
then the new collection $\mathcal N'$ of these intersection is still a local base of $0$.
One step further: The $x_n$'s, can be assumed to be linearly independent, for if
$$
x_m=c_1x_1+\cdot+c_kx_k,
$$
then $W_{x_1,1/\ell}\cap\cdots\cap W_{x_k,1/\ell}\subset W_{x_m,1/j}$,
if $\ell>j\big(\lvert c_1\rvert+\cdots+\lvert c_k\rvert\big)$.
From the above we derive that, if the weak$^*$ topology of $X$
is first countable, then there exists a linearly independent set
$\{x_n\}_{n\in\mathbb N}\subset X$, such that the finite  intersection of the open sets
$$
W_{x_n,1/k}, \quad k,n\in\mathbb N,
$$
form a local basis of $0$.
Now as $\dim X>\aleph_0$ we can find
$y\in X\smallsetminus\mathrm{span}\,\{x_n :n\in\mathbb N\}$. Using Hahn-Banach we can
construct a sequence $\{y_n^*\}_{n\in\mathbb N}\subset X^*$ satisfying
$$
y_n^*(y)=1 \quad\text{and}\quad y^*_n(x_j)=\frac{1}{n},\,\,\text{for $j=1,\ldots,n$.}
$$
Clearly, every open set of the form $W_{x_{i_1},1/k_1}\cap\cdots\cap W_{x_{i_n},1/k_n}$
contains all but finitely many terms of the sequence $\{y_n^*\}_{n\in\mathbb N}$, while
$W_{y,1}$ contains none of them. Hence
$$
W_{x_{i_1},1/k_1}\cap\cdots\cap W_{x_{i_n},1/k_n}\not\subseteq W_{y,1}.
$$
Thus,  the weak$^*$ topology on $X^*$ is not first countable if the dimension of $X$
is uncountable.
Corollary. Let $X$ be a normed space.
The weak$^*$ topology on $X^*$ is not first countable if the dimension of $X$,
as a linear space, is infinite.
Proof. If the dimension if $X$ is infinite, then the dimension of its
completion $\hat{X}$ is at least $\mathfrak{c}$ (the cardinal of the continuum),
while the weak$^*$
topologies of $X^*$ and $\hat{X}^*$ are identical.
