Closed subspace of dual space is the whole space if the intersection of kernels is 0

Question

Let $X$ be a Banach space, and $E \subset X^*$ a subspace of the dual $X^*$ that is closed in the weak-* topology. Show that if $\cap_{\lambda \in E} \ker(\lambda) = 0$, then $E = X^*$.

The analogous statement for finite dimensional spaces is easy to show, see for example here: Set of linear functionals span the dual space iff intersection of their kernels is $\{0\}$. In the infinite dimensional case, how to use closedness of $E$ to prove the statement? I was trying to use the fact that a neighborhood base of an functional $\lambda$ in the weak-* topology is of the form $$N(x_1, \cdots, x_n; \epsilon)=\{\ell \in X^* : |(\lambda-\ell)(x_i)| < \epsilon, i = 1, \cdots, n\}$$ to show $E$ is dense, but couldn't manage to complete the proof.

Answer 1

Suppose $\ell \in X^{*}$ but $\ell \notin E$. Then there exists a continuous linear functional on $(X^{*},weak^{*})$ which vanishes on $E$ but not at $\ell$. But continuous linear functional on $(X^{*},weak^{*})$ are just evaluations at some point of $X$. So there exist $x \in X$ such that $\ell (x) \neq 0$ but $\lambda (x)=0$ for all $\lambda \in E$. By hypothesis we get $x=0$ leading to a contradiction.