Separability of a product metric space

Your proof goes through. For countable products, make the analogous argument using the metric $$d(x,y)=\sum_{i=1}^\infty \frac{1}{2^i}\frac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)}$$ The summands are now bounded by $\frac{1}{2^i}$, so the sum always converges, and defines a metric on the product space.

Incidentally, if you take an even larger cardinality in your product, you're no longer guaranteed first countability, in which case you don't get a metric space. The product space in the pure topological category might still be separable, but it's no longer guaranteed. Steen and Seebach, Counterexamples in Topology, is a good place to look for general topology tidbits like this.

Two other thoughts about the answers so far:

1. Kevin in his answer gives you a metric that works for the countable product. But, you can actually use $\displaystyle{d(x,y)=\sum\frac{1}{2^i}d_i'(x,y)}$ where $d_i'$ is any metric that is compatible with $d_i$ and bounded by 1. So for example $d_i'(x,y)=\mbox{min}\{d_i(x,y),1\}$ would also work.

2. In the countable product space, the countable dense subset is no longer just the product of the dense subsets from each factor. Instead you have to fix a sequence of $\prod D_i$, say $\{x_i\}$ and take the subset of sequences that eventually agree with $\{x_i\}$. And this set is countable.

   This set is dense because any basic open set in the product $\prod X_i$ looks like $U=\prod U_i$ where the $U_i$ are open in $X_i$ and are not all of $X_i$ for only finitely many $i$, say up through $U_n$. So we can pick a sequence $\{s_i\}$ from our eventually constant set so that the $s_i$ is in $U_i$ for the first $n$ terms of the sequence. From then on let it agree with the $\{x_i\}$ above, and we will have $\{s_i\}\in U$.

   So for example in $\mathbb{R}^{\omega}$, the countable product of the reals, a dense countable subset is the set of all rational sequences that are eventually $0$.

I think it is not the answer for your question. It's better to see it as a comment after answering of Kevin Carlson. However I know there is a result from Pondiczery, Hewitt and Marczewski that (recently when I read a textbook I found):

If there are not more than $\mathfrak{c}$ ( which is the continuum), separable topological spaces, then their product is still separable.