In a nutshell, no, at least in the separable case. Let $F\subseteq E^*$ be not norm dense, and with $F$ (norm) separable. By Hahn-Banach there is $M\in E^{**}$ which is non-zero and annihilates $F$. Let $f\in E^*$ with $\langle M,f \rangle=1$.

I shall use Helly's Lemma (which I have failed to find an online reference for; it follows from e.g. the principle of local reflexivity) which says that if $N\subseteq

In a nutshell, no, at least in the separable case. Let $F\subseteq E^*$ be not norm dense, and with $F$ (norm-) separable. By Hahn-Banach there is $M\in E^{**}$ which is non-zero and annihilates $F$. Let $f_0\in E^*$ with $\langle M,f_0 \rangle=1$.

I shall use Helly's Lemma (which I have failed to find an online reference for; it follows from e.g. the principle of local reflexivity) which says that if $N\subseteq E^*$ is finite-dimensional and $M\in E^{**}$ then for $\epsilon>0$ we can find $x\in E$ with $\|x\|\leq \|M\|+\epsilon$ and $\langle M,f\rangle = \langle f,x\rangle$ for $f\in N$.

Let $(f_n)$ be a norm-dense sequence in $F$. For each $n$ there is hence $x_n\in E$ with $\|x_n\| \leq \|M\|+1/n$, with $\langle M,f_0\rangle = \langle f_0,x_n\rangle$ and with $\langle M,f_k\rangle = \langle f_k,x_n\rangle$ for $k\leq n$. Thus $(x_n)$ is not norm-null. I shall show that $\langle f,x_n \rangle \rightarrow 0$ for each $f\in F$.

(This follows as $(x_n)$ is bounded and $(f_n)$ is dense in $F$. To give the details, for $f\in F$ and $\delta>0$ there is $k$ with $\|f-f_k\|<\delta$ and so if $n\geq k$ then $|\langle f,x_n\rangle| $ $\leq |\langle f-f_k,x_n\rangle| + |\langle f_k,x_n\rangle| $ $= |\langle f-f_k,x_n\rangle| + |\langle M, f_k\rangle| $ $= |\langle f-f_k,x_n\rangle| $ $\leq \delta (\|M\|+1/k)$.)