Metrizability of the unit ball $B_{X^*}$. I am trying to prove the assertion:

If $X$ is a separable normed space, then the unit ball in $X^*$ with
  the weak* topology,  $(B_{X^*},\sigma(X^*,X))$, is metrizable.

Firstly, I took $D=(x_n)\subset X$, $\overline{D}=X$. If $\sigma(X^*,D)$ is the topology in $X^*$ with countable local basis for $0$:
$$U_n= \{x^* \in X^*: \sup\limits_{1\leq j\leq n}|x^*(x_j)|< \varepsilon_n \}$$
How can I prove that $(X^*, \sigma(X^*,D))$ is a Hausdorff TVS? If I could prove this, than I should conclude that $(X^*, \sigma(X^*,D))$ is metrizable.
Then I have another problem: How can I use Alaoglu's Theorem to show that $\sigma(X^*,D)=\sigma(X^*,X)$ in $B_{X^*}$ ?

Let $D=(x_n)$ dense in $X$ and let $\sigma(X^*,D)$ the locally convex topology in $X^*$ with countable local basis at $0\in X^*$ given by:
$$U_n=\{x^*\in X^*: \sup_{1\leq j\leq n}|x^*(x_j)|< 1/n \}$$
Then $(X^*,\sigma(X^*,D))$ is a Hausdorff TVS. In fact, if $x^*\neq 0$, as $x^*$ is continuous, there is $j_0$ such that $x^*(x_{j_0})\neq 0$, so we have $|x^*(x_{j_0})| \geq 1/n_0$. We have:
$$\forall m \geq n_0 \quad x^* \notin U_m $$
Now it is easy to see that $(x^* + U_{2n_0})$ is an open neighborhood of $x^*$ and $U_{2n_0} \cap (x^* + U_{2n_0})=\emptyset$. Hence, $(X^*,\sigma(X^*,D))$ is a Hausdorff TVS, so it is metrizable.
Clearly, $\sigma(X^*,D) \leq \sigma(X^*,X)$ and by Alaoglu's theorem, $B_{X^*}$ is $\sigma(X^*,X)$-compact. Then
$$ (B_{X^*},\sigma(X^*,X)) \stackrel{id}{\longrightarrow} (B_{X^*},\sigma(X^*,D)) $$
is an homeomorphism and $\sigma(X^*,X)=\sigma(X^*,D)$ in $B_{X^*}$, therefore $(B_{X^*},\sigma(X^*,X))$ is metrizable.
