If $E\subset X^{*}$ is bounded, then so is its weak* closure If $X$ is a Banach space and $E\subset X^{*}$ is norm-bounded, I've shown that its weak* closure is also norm-bounded using Alaoglu's theorem. But perhaps using Alaoglu's theorem is not necessary?
(I've shown that if $F\subset X$ is norm-bounded, then its weak closure is also norm-bounded...)
 A: I'll add details for completeness, avoiding the language of nets. It suffices to prove that a closed ball $B_R=\{\phi\in X^*: \|\phi\|\le R\}$ is weak*-closed; indeed, any bounded set is contained in such a ball. 
Take any $\phi \in X^*\setminus B_R$. By definition of the norm, there is a unit vector $u\in X$ such that $|\phi(u)|>R$.  Let $\epsilon =  |\phi(u)|-R$. The set 
$$U = \{\psi\in X^* : |\psi(u)-\phi(u)|<\epsilon\}$$
is weak* open, contains $\phi$, and is disjoint from $B_R$. Thus $B_R$ is weak* closed.
A: I see, Alaoglu's theorem is not needed: one can simply consider a weakly* convergent net in $E$ and show that its limit is also bounded. 
A: If you are comfortable with the language of nets, this is not too hard to prove.
Since $E$ is bounded, suppose all $f\in E$ satisfies $||f|| \le C$ for some constant $C$. Let $f'$ be in the weak$^*$ closure of $E$. Then these exists a net $\langle f_{\alpha}\rangle$ such that $f_{\alpha}\to f'$ in a weak$^*$ manner, i.e. $f_{\alpha}(x)\to f'(x)$ for all $x\in X$.
For any fixed $x$ satisfying $||x|| = 1$, we have $||f_{\alpha}(x)||\le ||f_{\alpha}|| \le C$. Since $f_{\alpha}(x)\to f'(x)$, we have $||f'(x)||\le C$. By definition of the norm of a functional, $||f'||\le C$. This shows that the weak$^*$ closure of $E$ is also bounded (by the same constant!).
Remark: OP stated that he/she managed to prove that if $F\subseteq X$ is norm-bonded, then so is the weak-closure. I would have thought that the proof is similar and in fact harder in this latter case (having to use the isometry of $X$ into $X^{**}$).
