Given $K^w$ equipped with product topology is an infinite product of countably infinite discrete space $K$ . Show that $K^w$ is second countable.
My Progress: Since the product topology means there exists a basis $\cup U_i$ such that $U_i$ is open in $K_i$ for each $i$ and $U_i\neq K_i$ at only finitely many values of $i$. Thus the basis is countable since we only care about the difference part between the basis and the product topology, and there are only finitely many $U_i$, which means our basis is countable.
I don't feel quite satisfied with this naive proof, and I think there must be some mistakes in it. Can someone please help correct/modify it?