Is it possible to show $\mathbb{R}^\omega$ is not metrizable without using the first countable definition and sequence lemma?

Question

I encountered a question about proving that $\mathbb{R}^\omega$, which is the countably infinite product of $\mathbb{R}$ under the product topology, is not metrizable. I have seen many solutions here by using the sequence lemma and the first countable argument, which is great. However, are there any ways to prove that by just using basic definitions of product topologies and metric spaces? I was trying to bring up some contradiction about violating the definition of being a metric but it seems nothing is wrong with that...

Answer 1

This is false. A countable product of metrizable sets is metrizable (w.r.t. the product topology). In your case, $\Bbb{R}^\omega \cong \Bbb{R}^\Bbb{N}$ is for example metrizable with metric

$$d((x_n)_n, (y_n)_n):= \sup_{n \in \Bbb{N}}\frac{1}{n} (|x_n-y_n| \land 1)$$