Let $A$ be a complex normed vector space, $\varphi\in A^*$ and $\epsilon>0$. Of course $B_\epsilon(\varphi)$ is open in the norm topology, but what about in the weak-* topology? For each $a\in A$, the evaluation $J_a$ given by $\varphi\mapsto\varphi(a):A^*\to\mathbb{C}$ is continuous in the weak-* topology by definition. We know that $$\overline{B}_\epsilon(\varphi)=\bigcap_{\|a\|\leq1}J_a^{-1}(\overline{B}_\epsilon(\varphi(a)))$$ is weak-* closed, but this approach won't work for open balls because the intersection is infinite.
EDIT: Does the proof given by the OP here assume open balls to be weak-* open?
Here's an example of an argument. Non-empty weak* open sets always contain a translate of a finite-codimension sub-space. In particular for infinite dimensional spaces they may never be bounded.
Recall a sub-basis of the topology is given by sets of the form $\Psi^{-1}(U)$ with $U$ open in $\Bbb K$ and $\Psi$ a weak*-continuous functional. If $x$ is such that $x\in \Psi^{-1}(U)$ (ie the set is not empty) then clearly $x+\ker(\Psi)\subseteq \Psi^{-1}(U)$. Since $\Psi$ is a linear functional you have that $\ker(\Psi)$ has co-dimension one.
Now suppose you have a finite intersection $$\bigcap_{i=1}^n \Psi_i^{-1}(U_i)$$ if this is not empty there is some $x$ in it and then clearly $x+\bigcap_{i=1}^n \ker(\Psi_i)$ also is contained in this intersection. This intersection of kernels has co-dimension $n$, since its an intersection of $n$ co-dimension $1$ spaces.
This argument works for any topology induced by a space of linear functionals.