Summary
I believe I've written a geometric group theory flavoured proof with a mistake in it, but I'm struggling to see why it might be wrong. I haven't found a counter example, but it also feels too good to be true. Can someone please verify my proof?
Background
I'm currently trying to understand generators of groups from group actions on topological spaces. The case of importance is group actions on simplicial complexes, and I've proven the desired result in a very convincing manner. Namely, I showed that if $D$ is a subcomplex of a connected simplicial complex $C$, $G$ acts by simplicial automorphisms, and $GD = C$, then $S = \{g\in G: gD\cap D \neq \varnothing\}$ generates $G$. The fact that $C$ is a simplicial complex was crucial in my proof.
However, when I attempted to generalise the result, I ended up obtaining a much more general result with a much simpler proof. For this reason I suspect there is a mistake in my proof. Can someone please correct me if I've made an assumption somewhere?
Statement and Proof
Claim: Let $X$ be a connected topological space, and $G$ a group acting on $X$. Suppose $D$ is an open subset of $X$ such that $GD = X$. Then $S = \{g\in G: gD\cap D \neq \varnothing\}$ generates $G$.
Proof: Let $A = \langle S\rangle$. We first show that $(G\setminus A)D$ is disjoint from $AD$. Let $g \in G$, and suppose that $gD \cap AD \neq \varnothing$. Then there exists $a \in A$ such that $gD \cap aD \neq \varnothing$, so $a^{-1}g D \cap D \neq \varnothing$. By definition, this means $a^{-1}g \in S$, so in particular $g \in aS \subset A$. Since this shows that $g$ must have been in $A$ to begin with, $(G\setminus A)D$ is disjoint from $AD$.
But recall that $D$ was open. A group action is a map $G \to \text{Aut}(X)$, so $gD$ is open for every $g \in G$. $AD$ and $(G\setminus A)D$ are then unions of open sets, and are themselves open. Moreover, $AD \cup (G\setminus A)D = X$. Since $X$ is connected, this forces either $AD$ or $(G\setminus A)D$ to be empty. The former is clearly nonempty, so $(G\setminus A)D = \varnothing$. This forces $G = A$ as required.
I believe I've seen a similar result in the past and I also felt that it had additional properties, but I can't say for sure. Thank you!