Here is my proof. Let $C$ be the pre-polar of D:

$C$ = {$x: sup_{f \in D} |f(x)| \leq 1$} 

By applying Bishop-Phelps to $C$, we get that $A(C):=${$f \in X^*:\exists x_0 \in C$ s.t $f(x_0)=sup_{x \in C} f(x) $}
is dense in $D$.
Fix some $f_0 \in \partial D$. We may choose a sequence $(g_n) \subset D \cap A(C) $ s.t $g_n \to f_0$. We may normalize $g_n$ (i.e, multiply by appropriate scalars, while keeping the same name) s.t $g_n \to f_0$ and now $g_n \in \partial D \cap A(C) $ (as $A(C)$ is invariant under positive scalar multiplication).

Now, i claim that if $g \in A(C) \cap \partial D$, then $g$ is a $w^*$-support point for $D$. Indeed, By Hahn-Banach, there exists some $\phi_0 \in X^{**}$ s.t $ \phi_0(g)=max \phi_0(D) = 1$. 
Let $C^{**} := $ {$ \phi \in X^{**}:sup_{f \in D} |\phi(f)| \leq1 $}. So, we have $\phi_0 \in C^{**}$, $C = C^{**} \cap X $, and by Goldstein, $C$ is $w^*-$dense in $C^{**}$. Therefore, we may choose $(x_n) \subset C$ s.t $x_n \to \phi_0$ ($w^*$ convergence). Thus,
$ x_n(g) = g(x_n) \to \phi_0(g) = 1$. But, since $g \in A(C), \exists x_0 \in C$ s.t $g(x_0)=sup_{x \in C} g(x) = 1$. But by the definition of $C$, $|x_0(D)| \leq 1$. Hence $g(x_0)=1=sup_{f \in D}f(x_0)$, i.e, $g$ is a $w^*$-support point for $D$.

We fixed $ f_0 \in \partial D$ and found a sequence $(g_n)$ of $w^*$-support points for $D$ which convegre to $f_0$. Hence the set of $w^*$-support points for $D$ is dense in $\partial D$.