Unit ball of a Separable Banach Spaces is metrizable Please help me to understand why the unit ball of a separable banach space is metrizable, when it is given the induced topology from the weak topology.  Specifically, my problem is I think I'd be able to do it if I knew the dual was separable, but not with the current given information. (If I had this additional information, I'd use the infinite sum metric tricks that one uses all too often.)
The underlying field can be real or complex.
 A: I assume you mean metrizability of the unit ball in the dual space. Since you don't ask about compactness, we can cheat and assume Alaoglu's theorem that the unit ball is compact in the weak*-topology. Since a continuous injective map from a compact space to a Hausdorff space is a homeomorphism onto its image, it suffices to find a continuous and injective map into some metrizable space.
Choose a dense sequence $\langle x_n : n \in \mathbb{N}\rangle$ in the unit sphere of $X$, this exists by separability of $X$. Let $B'$ be the unit ball in the dual space. By the definition of the weak*-topology we have a continuous map $f_n \colon B' \to D$ given by $f_n(\varphi) = \varphi(x_n)$ where $D = \{z \in \mathbb{C} : |z| \leq 1\}$ is the closed unit disc in $\mathbb{C}$. These assemble to a continuous map $f \colon B' \to D^{\mathbb{N}}$ given by 
$$f(\varphi) = \langle f_n(\varphi) : n \in \mathbb{N}\rangle = \langle \varphi(x_n) : n \in \mathbb{N}\rangle$$ by the definition of the product topology on $D^\mathbb{N}$. I claim that this map is injective. Indeed, if $\varphi_1 \neq \varphi_2$ then there is $x \in X$ such that $\varphi_1(x) \neq \varphi_2(x)$. Normalizing $x$ we can assume hat $\|x\| = 1$. By choosing $x_n$ close enough to $x$ we will have $\varphi_1(x_n) \neq \varphi_2(x_n)$. This implies that $f(\varphi_1) \neq f(\varphi_2)$ and we're done.
To be a little more explicit, the standard metric on $D^\mathbb{N}$ is $d(a,b) = \sum_{n=1}^\infty 2^{-n} |a_n - b_n|$ and since $f$ is injective and continuous, $\delta(\varphi_1,\varphi_2) = d(f(\varphi_1),f(\varphi_2))$ is a continuous metric on $B'$ with the weak*-topology and this shows metrizability of $B'$.
We assumed Alaoglu's theorem. If we don't do that there are two more things to do: 


*

*show that convergence with respect to $\delta$ is equivalent to weak*-convergence.

*show that $B'$ is compact with respect to $\delta$.

