On the product of $\mathfrak c$-many separable spaces

I already figured out how to show that the countable product of separable topological spaces is separable, but I'm out of ideas when the index set has cardinality of $\mathfrak c$. My textbook says it is possible but gives no references. Any suggestions how to prove this statement?

In a less general setting, I would also be interested to see how a dense countable set is constructed to $\mathbb{R}^{\mathbb{R}}$. Thanks in advance.

This is a special case of Hewitt-Marczewski-Pondiczery theorem, see e.g. Theorem 2.3.15 in Engelking's General Topology:

If $d(X_s)\leq \alpha\geq\aleph_0$ for every $s\in S$ and $|S|\leq 2^\alpha$, then $d(\prod X_s)\leq\alpha$.

The $d(X)$ denotes the density of the topological space $X$, which is defined as $$d(X)=\min\{|D|; D\text{ is a dense subset of }X\}+\aleph_0.$$ I.e., $d(X)$ is the smallest cardinality of a dense subset, but if there is a finite dense subset, we put $d(X)=\aleph_0$.

This means that a topological space is separable if and only if $d(X)=\aleph_0$.

Some further references are given at Planetmath. Wikipedia article on separable space mentions Theorem 16.4c in Willard's General Topology as a reference for the special case you're asking about.

A proof of this theorem can be found at in this post from Ask a Topologist. (The post was written by Henno Brandsma.)

This theorem can be used to show that there is an independent family on $\mathbb N$ of cadinality $\mathfrak c$; see Stephan Geschke's MO post and paper.

For $\mathbb{R}^\mathbb{R}$, try viewing it as the set of all functions from $\mathbb{R}$ to $\mathbb{R}$. Then the set of polynomials with rational coefficients is a countable dense subset. (Show that every nonempty open set contains such a polynomial. It might help to first show that every nonempty open set contains a polynomial with real coefficients.)