I just wanted to point out how Andreas' proof above relates to the separability of products of $2^{\aleph_0}$ separable spaces:

Let $\lambda<(2^{\aleph_0})^+$.
Suppose $\langle X_\alpha:\alpha<\lambda\rangle$ is a family of separable spaces.  For each $\alpha<\lambda$ let $\langle x_\alpha^n:n\in\omega\rangle$ enumerate a dense subset of $X_\alpha$.
Let $F:\lambda\times\omega\to\omega$ be as in Andreas' proof.
Now the set of sequences of the form
$\langle x_{\alpha}^{F(\alpha,n)}:{\alpha<\lambda}\rangle$, $n\in\omega$, is dense in $\prod_{\alpha<\lambda}X_\alpha$.