If for every $e\in E\setminus\{0\},$ there exists $e^*\in A$ such that $e^*(e)\neq 0,$ then $A$ is weak-star dense in $E^*?$

Question

Suppose that $E$ is a Banach space and $E^*$ is a dual space of $E,$ that is, $E^*$ contains all bounded linear functionals on $E.$
Let $A$ be a closed subspace of $E^*.$
Equip $E^*$ with weak$^*$-topology.

Question: To show that $A$ is weak-star dense in $E^*,$ is it sufficed to show that for every $e\in E\setminus\{0\},$ there exists $e^*\in A$ such that $e^*(e)\neq 0?$

My lecturer used the above technique to conclude that $A$ is weak-star dense in $E^*.$
If it is true, may I know how to prove it?

Answer 1

This is immediate from Hahn - Banach Theorem. If $A$ is not weak$^{*}$ dense the there is a continuous linear functional which is $0$ on $A$ but not identically $0$. Now use the fact that dual of $(E^{*},$ weak$^{*})$ is nothing but $E$.