Consequence of metrizability proof - disregard, the question is an error

In Marian Fabian et al's Functional Analysis and Infinite-Dimensional Geometry, Proposition 3.22 states/proves that if $X$ is a separable Banach space, then the (closed) unit ball, $B_{X^{*}}$ of $X^{*}$, endowed with the weak-$*$ topology of $X^{*}$ is metrizable.

They do this by defining a metric $d$ on $X^{*}$ using a dense subset of the unit sphere of $X$, and then showing the identity map $I:(B_{X^{*}},weak-*)\to (B_{X^{*}}, d)$ is a homeomorphism.

The importance in restricting to the unit balls comes from the use of Alaoglu's theorem which states that $B_{X^{*}}$ is compact in the weak-$*$ topology. This avoids the problem of having to show that $I^{-1}$ is continuous, as $(B_{X^{*}},weak-*)$ is compact, and $(B_{X^{*}}, d)$ is Hausdorff.

Note: the comment below is incorrect and was a complete misread of the statement. Please disregard the question.

After the proof is completed, it is concluded as a remark that the spaces $(X^{*},weak-*)$ and $(X^{*},d)$ are in fact homeomorphic and thus the weak-$*$ topology of $X^{*}$ is metrizable.

I have a problem proving this Corollary. Continuity on the unit ball can easily be extended to the whole space since scalar multiplication (and thus normalization) is a homeomorphism in a topological vector space, but as far as I know we have no such guarantee in a metric space. Can anyone suggest how I can see the truth behind this consequence?

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It is simply wrong that $X^\ast$ is metrizable in the weak$^\ast$-topology. The weak$^\ast$-topology on $X^\ast$ is not first countable if $X$ is infinite-dimensional. – t.b. Apr 12 '12 at 19:06
I just did a double-check and I completely misread the consequence. Quite embarassing, actually. Thanks for pointing this out. – Kevin Apr 12 '12 at 19:15