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 $B_{X^*}$ ?

I don't immediately see if your approach leads to a solution (in particular I think you need "normed" for this to be true). 
This is the way I would do it: assume that $\|\phi\|\leq c$ for all $\phi\in K$. Let $D=(x_n)$ be dense in the unit ball of $X$. Now define
$$
d(\phi,\psi)=\sum_{k=1}^\infty2^{-k}|\phi(x_k)-\psi(x_k)|.
$$
It is straighforward to check that this is a distance. 
Assume $d(\phi_n,0)\to0$. Fix $x\in X$, $\varepsilon>0$. Then there exists $k$ with $\|x-x_k\|<\varepsilon$. Then
$$
|\phi_n(x)|\leq\|\phi_n\|\,\|x-x_k\|+|\phi_n(x_k)|\leq c\,\varepsilon+2^k\,d(\phi_n,0).
$$
Then
$$
\limsup_n|\phi_n(x)|\leq c\varepsilon.
$$
As $\varepsilon$ was arbitrary, we get $\limsup_n|\phi_n(x)|=0$, i.e. $\phi_n(x)\to0$. So $\phi_n\to0$ in the $\sigma(X^*,X)$ topology. 
Conversely, if $\phi_n\to0$ weak*, then $\phi_n(x)\to0$ for all $x\in X$. Fix $\varepsilon>0$. Let $k\in\mathbb N$ with $2^{-k}<\varepsilon/(2c)$. Then there exists $n_0$ such that $|\phi_n(x_j)|<\varepsilon/2$ for $j=1,\ldots,k$ for all $n\geq n_0$. Then, for such $n$, 
$$
d(\phi_n,0)=\sum_{j=1}^k2^{-j}|\phi_n(x_j)|+\sum_{j=k+1}^\infty2^{-j}|\phi_n(x_j)|
\leq\frac\varepsilon2\,\sum_{j=1}^k2^{-j}+c\sum_{j=k+1}^\infty2^{-j}\leq\frac\varepsilon2+\frac\varepsilon2=\varepsilon.
$$
