We know that $2^{\mathfrak c^+}$ ($\mathfrak c =2^\omega=|\mathcal P (\omega)|$) is not separable by the following argument:
Suppose $D$ is countable dense in $2^{\mathfrak c^+}$. For each $\alpha<\mathfrak c ^+$ we have $D_\alpha=\pi_\alpha ^{-1} \{0\} \cap D\neq\varnothing $. If $\alpha\neq\beta<\mathfrak c ^+$ then we have $D_\alpha \cap \pi_\beta ^{-1} \{1\} \neq\varnothing $ but $D_\beta \cap \pi_\beta ^{-1} \{1\}=\varnothing $, whence $D_\alpha\neq D_\beta$. So we have $\mathfrak c^+>\mathfrak c$ different subsets of our countable set $D$, which is a contradiction.
The same argument does not necessarily work for $2^{2^{\omega_1}}$: we would get $2^{\omega_1}$ many different subsets of a countable set, but this is not necessarily a contradiction(!)- there exist models of set theory (for instance any model in which Martin's Axiom holds), where you can have $2^\omega=2^{\omega_1}$. My question: In any of these models could we actually get $2^{2^{\omega_1}}$ separable?
Thank you!