That is, I want to cover $X^*$ (X is Banach space) with a family $\{U_{\alpha}\}$, where $diam(U_{\alpha})<\epsilon$ and each $U_{\alpha}$ is weak* open.
I expect, that not every open ball is weak* open, otherwise norm and weak* topologies would coincide (as in case of the reflexive spaces). But still, maybe there are enough $\epsilon$-balls which are weak* open to achive such cover? Or we can count for a cover with less regular sets?
If answer to the question is affirmative, what's the proof. And if not, when it is true? Namely, we have a metric space $X$, and consider it with some topology $\tau$. What properties does the $(X,\tau)$ need to posses to be sure, such cover exists? If i'm not mistaken, one trivial answer is, when $X$ is finite dimensional and $\tau$ is generated by any norm on $X$. Another one was mentioned above.
Thanks for any help.