The theorem states that if $F$ is regular and has a countable basis, then it is metrizable. In Munkres' proof of this theorem, he gives a function (homeomorphism) $F:X \rightarrow [0,1]^\omega$ that embeds it into a subspace of $[0,1]^\omega$ with the product topology. This subspace is metrizable (it can be easily seen), and thus the theorem is proven. However, the author also provides an alternative proof, where using the same function $F$, he embeds $X$ into a subspace of $[0,1]^\omega$ with the uniform topology! My question is, if this subspace of $[0,1]^\omega$ is $Z$, then this proves that $Z$ under the product topology is homeomorphic to $X$, which is in turn homeomorphic to $Z$ in the uniform topology. Thus $Z$ under one topology is homeomorphic to $Z$ under the other one. But this is absurd, isn't it?
I greatly appreciate your help!
You got fooled: It's not the same function $F$. To get continuity of the second map (into $[0,1]^\omega$ in the uniform topology), for each $n\in\Bbb N$ he scales the function $f_n$ by a factor of $1/n$. So the image is a compressed version of the image of the first mapping.
Comment: If you consider the Hilbert Cube $\prod\limits_{n=1}^\infty [0,1/n]\subset [0,1]^\omega$, the product and uniform topologies are indeed the same, so then your alleged absurdity disappears. :)