for a 'non-reflexive' space, the weak star topology is strictly coarser than the weak topology on the dual space

Question

I'm trying t0 work out the fine details of a claim in "Functional analysis, Sobolev spaces and PDE's" by H. Brezis. Let $J: \begin{cases} E\longrightarrow E^{**} & \\ x \longmapsto J(x) & \end{cases}$, with $J(x)\left ( f \right )= f(x), \forall f \in E^*$ and suppose that $J$ is not surjective (for this text by definition meaning that the normed linear space $E$ is non-reflexive), then the weak$^*$-topology is strictly coarser than the weak topology on $E^*$. First, the author has shown, using the separating hyperplane theorem for two disjoint convex subsets of a Euclidean space: $Lemma.$ a hyperplane in $E^*$ that is weak$^*$ - closed, has the form $\left \{ f \in E^* | J(x_0)\left ( f \right )= \alpha \right \}$, for some $x_0 \in E, x_0 \neq 0$ and some $\alpha \in \mathbb{R}$. I get all that. Then, he says, if $\xi \in E^{**}$, $\xi $ not in the image of $J$, consider $H=\left \{ f \in E^* | \xi(f)=0 \right \}$. Obviously, $H$ is a hyperplane in $E^*$ that is weakly closed. But it isn't weak$^*$-closed because of $Lemma$. No further explanation. So, I was thinking: because of the $Lemma$, one must have that $H=\left \{ f \in E^* | J(x_0)(f)=\alpha \right \}$. Now, $H$ is linear but $\left \{ f \in E^* | J(x_0)(f)=\alpha \right \}$ has a non-zero element if $\alpha \neq 0$ (take $f_0 \in E^*$ with $f_0(x_0)= \lVert x_0\rVert^2$, then $\dfrac{\alpha f_0}{\lVert x_0\rVert^2}$ will do), hence $\left \{ f \in E^* | J(x_0)(f)=\alpha \right \}$ can never be linear, so we allready have $\alpha = 0$. Where do I go from here? Maybe I'm missing something and completely barking up the wrong tree. Any ideas? To avoid further spam, many thanks in advance.