Is $2^{\frak{c}}$ separable? Let $2 = \{0,1\}$ be endowed with the discrete topology. My hunch was the following:
Let $e_\mathbb{N}: \mathbb{N} \to 2^{2^{\mathbb{N}}}$ be the "evaluation map", that is, it is given by $$e_\mathbb{N}(n): 2^\mathbb{N}\ni f\mapsto f(n).$$
Then I hoped that $\text{im}(e_\mathbb{N})\subseteq 2^{2^{\mathbb{N}}}$ is dense, but I fear that if $\underline{1}\in 2^\mathbb{N}$ is the constant $1$-sequence, then $\pi^{-1}_{\underline{1}}(\{0\}) \cap \text{im}(e_\mathbb{N}) =\emptyset$.
So am I just misremembering something, or is the mistake elsewhere? Maybe $2^{2^{\mathbb{N}}}$ is not separable at all?
 A: You need to take the set of all (finite) Boolean combinations of your elements $e_\mathbb{N}(n)\in 2^{2^{\mathbb{N}}}$ (here I mean "Boolean combination" in the usual sense when you think of elements of $2^{2^{\mathbb{N}}}$ as subsets of $2^{\mathbb{N}}$).  These Boolean combinations will still form a countable set, and actually are dense.
Indeed, given a finite subset $S\subset 2^{\mathbb{N}}$, you can find a finite subset of $A\subset\mathbb{N}$ that distinguishes all the elements of $S$.  Any function $2^A\to 2$ can then be written as a Boolean combination of the functions given by restricting $e_\mathbb{N}(n)$ for $n\in A$, and in particular this means that for any function $f:S\to 2$ there is a Boolean combination of the $e_\mathbb{N}(n)$ that agrees with $f$ on $S$.
A: To get a countable dense subset of $2^\mathbb R,$ take the set of all step functions with finitely many jumps at rational points on the line. By a similar argument, any product of continuum many separable spaces is separable. (Separability of the product space $Q^Q$ where $Q=[0,1]$ is part (a) of exercise N in Chapter 3 of John L. Kelley's General Topology, which is available at the Internet Archive.) "Continuum many" is best possible here, seeing as the maximum possible cardinality of a separable Hausdorff space is $2^{2^{\aleph_0}}$.
