Let $G$ be a discrete group which acts continuously on a Stonean space $\Omega$. Consider the map $f\colon \Omega\to \{0,1\}^G$ sending $x\in \Omega$ to $\chi_{G_x}$, where $\chi_{G_x}$ denotes the characteristic function of the stabilizer $G_x$.

Why is $f$ continuous?

We can restrict to the subbasis-sets of the product topology and for $g\in G$ fixed, it is enough to consider the preimages of $\{0\}\times \{0,1\}^{G\setminus \{g\}}$ and $\{1\}\times \{0,1\}^{G\setminus \{g\}}$ under $f$, which must be open in $\Omega$. That $f^{-1}(\{0\}\times \{0,1\}^{G\setminus \{g\}}) $ is open in $\Omega$ can be seen using that in the Hausdorff space $\Omega$, the diagonal $\{(x,x)\in \Omega\times \Omega\mid x\in \Omega\}$ is closed in $\Omega\times \Omega$, but without using that $\Omega$ is Stonean.

My question therefore reduces to: Why is $ f^{-1}(\{1\}\times \{0,1\}^{G\setminus \{g\}}) $ open in $\Omega$ ? I think to prove this, it must be used that $\Omega$ is Stonean and probably it helps to know that in a Stonean space, the closure of disjoint open sets is again disjoint.

The background of my question is the following: I am currently reading the paper https://arxiv.org/pdf/1509.01870.pdf and that $f$ is continuous is used in the proof of Theorem 4.1, direction $<=$.