From
http://www.proofwiki.org/wiki/Banach-Alaoglu_Theorem
there were just a few more steps:
- Let $l(g_i) = \lim_{n_k} \langle x_{n_k} , g_i \rangle = \langle x_\infty, g_i \rangle$. (We can obtain $x_\infty$ by including the standard basis in the dense $g_i$).
- Extend to all $g \in l^2$ via $l(g) = \lim_i l(g_i)$. This makes $l$ a linear functional, and can be bounded based on $x_{n_k}$. $|l(g)| \leq |l(g)-l(g_i)| + |l(g_i) - l_{n_k}(g_i)| + |l_{n_k}(g_i)|$.
- Riesz representation and uniqueness shows that $l(x) = \langle x_\infty, x \rangle$.
Then we have the weak convergence as desired.
A normed space has the property (that all metric spaces have) that $X$ is separable then all subsets are separable too in their subspace topology. So $X^\ast$ (norm)-separable implies that $B_{X^\ast}$ is separable. The reverse also holds in all normed spaces $Y$ by "scaling": if $x \neq 0$ then choose $\alpha = \frac{1}{\|x_n\|}$ so that $\|\alpha x\| =1$. If we then (by separability of the unit ball) find a sequence $d_n$ on the ball $B_Y$ that converges to $\alpha x$, and also a sequence of rationals $q_n \to \frac{1}{\alpha}$. Then $q_n d_n \to \frac{1}{\alpha} (\alpha x) = x$. This essentially shows that if $D$ is countable and dense in $B_Y$ then $\{qd: q \in \mathbb{Q}, d \in D\}$ is (countable and) dense in $Y$. So $B_Y$ separable iff $Y$ separable. So we lose nothing by using the unit ball (here the sphere really). And the Hahn-Banach theorem which links things in $X$ to $X^\ast$, gives us functionals on $B_{X^\ast}$ anyway.
Using the unit sphere makes things easier, because you know the norm of all dense elements, namely 1, which allows for the choice of the $x_n$ (otherwise we'd need to scale there too which makes for a more messy proof). We have to prove something on $X$, so we can find points $x_n$ on which $f_n$ is relatively large: we know that $\|f_n\| = \sup \{|f_n(x)|: x \in B_X \}$, so we can find $x_n \in B_X$ such that $|f(x_n)|$ is as close to $1$ as we like. Here more than $\frac{1}{2}$ is sufficient.
As above a countable dense set in $B_X$ is enough to get one on $X$, using the span with rational coefficients. So try that for the $D = \{x_n: n \in \mathbb{N}\}$ we now have: taking finite sums from $D$ with rational coefficients we can approximate all members of the linear span of $D$ (just approximate real coefficients in $\mathbb{R}$ by members of $\mathbb{Q}$; we use that the field is separable). This $\operatorname{span}_{\mathbb{Q}}(D)$ is still countable (standard set theory argument: finite products of countable sets are countable and a union of countably many countable sets is countable). So $Y = \overline{\operatorname{span}(D)}$ has a countable dense set $\operatorname{span}_{\mathbb{Q}}(D)$. So we'd be done if $Y =X$. So assume it's not.
Then, Hahn-Banach allows us to find a functional $f$ (back to $B_{X^\ast}$ where we know something about the $f_n$) that has norm $1$ and is $0$ on $Y$. (In particular $f$ is such, that it is $0$ on the set $D$ where the $f_n$ are chosen to be large, and so the $f$ is very "different" from the $f_n$ and so we cannot approximate it in norm by the $f_n$.)
So we know $f(x_n) = 0$ (as $x_n \in Y$!) and the final part of the proof shows that if $f_n$ is chosen to be close to $f$, the point $x_n$ where $f_n$ is large shows that $f$ should also be at least $\frac{1}{4}$ in that point as well.
This gives the needed contradiction.
Best Answer
It follows from $|x^\ast(x)| \le \|x^\ast\|\cdot|x|$ for all $x^\ast \in \mathscr{X}^\ast, x \in \mathscr{X}$ (by definition of the norm on the dual):
If we take the $(x_n) \in \text{ball } \mathscr{X}$. We want the $\tau$ to be well-defined. And as $x^\ast \in \text{ball }\mathscr{X}^\ast$, $\|x^\ast\| \le 1$ and so for any $x_n \in \mathscr{X}$, $|x^\ast(x_n)| (= |\langle x_n,x^\ast \rangle|) \le \|x_n\|$. But we want them all to be in $\prod_n D_n$ (just because we can..), so we want $|\langle x_n, x^\ast \rangle | \le 1$, and so we choose all $x_n$ in $\text{ball } \mathscr{X}$.
This can be done as $\mathscr{X}$ being separable implies that all its subspaces are separable too (this holds in any metric space, in fact). As a matter of fact, in any Banach space, the fact that the ball of the space is separable implies the whole space is, too: just take all $qx$ where $x \in D$ (a countable dense set of the ball) and $q$ rational which is a countable dense subset of the whole space. This sketches that "$\text{ball }\mathscr{X}$ is separable" iff "$\mathscr{X}$ is separable". So we can use whichever is convenient.
For the second point, we use this classical fact from general topology:
In the proof we suppose we have $\tau(x^\ast) = \tau(y^\ast)$, so $\langle x_n, x^\ast \rangle = \langle x_n, y^\ast \rangle$ for all $n$, or $x^\ast - y^\ast$ is a function from $\text{ball } \mathscr{X}$ to $\mathbb{R}$ that agrees with the $0$-functional on a dense subset of the ball. So the above fact gives that it equals $0$ on all of $\text{ball } \mathscr{X}$, as claimed. As an alternative: use that for any $x \in \text{ball }\mathscr{X}$ there is sequence $x_m$ from the countable dense set such that $x_m \to x$ and use (sequential) continuity of $x^\ast -y^\ast$ to see that its value at $x$ is also $0$).
We indeed don't need $X$ to be compact. We show $\tau$ from the weak$^\ast$ ball $B$ to $X$ is 1-1 and continuous. Then $\tau[B] \simeq B$ as $B$ is compact and $X$ is metrisable (so Hausdorff). And $\tau[B]$ is metrisable, as a subspace of $X$. But we could just have used $X = \mathbb{F}^\mathbb{N}$ for that as well.
This proof is similar to the proof that $\text{ball }\mathscr{X}^\ast$ is compact: there we use all $x \in \text{ball }\mathscr{X}$ and define $\tau': \text{ball }\mathscr{X}^\ast \to D^{|X|}= \prod_{x \in X} D_x$ as $\tau'(x^\ast) = (\langle x, x^\ast \rangle)_{x \in X}$, where $D_x= D$ is the unit disk in $\mathbb{F}$. Here we need the $\le 1$ condition to get all the components in $D$, and have a product of compact spaces. $1-1$-ness is almost by definition here (if functionals agree on all points of the ball they're equal everywhere). Standard theorems in topology gives us that $\tau'$ is an embedding, and then what remains to be shown is that $\tau'[\text{ball }\mathscr{X}]$ is closed in this compact product..
The idea of the asked about proof is to "get away with" using only countably many points from $\text{ball }\mathscr{X}$ instead of all of them, and reduce the product to a countable (hence metrisable) one, but we can only do that if we can approximate all the points of the ball, i.e. we need a countable dense subset of it.
I think this proof was conceived as an expansion on the above idea, so it makes sense to start with a dense set in $\text{ball }\mathscr{X}$, even as for the metrisability it's not strictly needed.