What I did was considering the collection of linear bounded functionals $\left \{ Jx ; x\in E \right \}$ where $J$ is the canonical injection from $E$ to its bidual . ($\left \langle Jx,f \right \rangle=\left \langle f,x \right \rangle$)
For all $x$ in $E$ , $Jx$ are continuous in the weak* topology (by its definition) and $K$ is weakly*-compact , then $Jx(K)$ is bounded in $\mathbb{R}$.
Then, $\underset{x\in E}{sup} \left | \left \langle Jx,f \right \rangle \right | < \infty $
(for all $f$ in $K$)
Then by uniform boundedness principle , $\underset{x\in E}{sup} \left \| Jx \right \|< \infty$
I'm stuck here , what I want to show , is that $\left \| f \right \|< \infty$
(I know that $\left \| Jx \right \|=\left \| x \right \|$ , is it useful ? )
It seems you are making a mistake in the first line with a formula:
If $K$ ist weak$^{\ast}$-compact and $Jx$ is weak$^{\ast}$-continuous for each $x \in E$, then for every $x \in E$ we have
$$ \sup_{f \in K} \vert \langle Jx, f \rangle \vert < \infty $$
The $\sup$ does not go over $x \in E$, but over $f \in K$.
Can you take it from here?