Can someone help clarify a point in the proof of Theorem $1.55$ in John Meier's "Groups, Graphs and Trees?"
Theorem $1.55.:$ Let $G$ act on a connected graph $\Gamma$ with fundamental domain $\mathcal{F}$. Then the set of elements
$S = \{g \in G| g\neq e \text{ and } g\cdot \mathcal{F} \cap \mathcal{F} \neq \emptyset \}$
is a generating set for $G$.
Proof: Let $g\in G$ be an arbitrary element and let $v$ be a vertex in the fundamental domain $\mathcal{F}$. Choose a path $p$ joining $v$ to $g\cdot v$ and $\{\mathcal{F}, g_{1}\cdot \mathcal{F}, g_{2}\cdot \mathcal{F}, \ldots, g_{n}\cdot \mathcal{F} = g\cdot \mathcal{F} \}$ be a finite sequence of images of the fundamental domain such that
the entire path $p$ is contained in $\cup g_{i}\cdot \mathcal{F}$, and
$ g_{i}\cdot \mathcal{F} \cap g_{i+1}\cdot \mathcal{F} \neq \emptyset$, and $g_{0}$ is defined to be the identity.
Since $\mathcal{F} \cap g_{1}\cdot \mathcal{F} \neq \emptyset$, $g_{1}\in S$ by definition. Similarly $ g_{1}\cdot \mathcal{F} \cap g_{2}\cdot \mathcal{F} \neq \emptyset$ implies $\mathcal{F} \cap g_{1}^{-1}g_{2}\cdot \mathcal{F} \neq \emptyset$, thus $ g_{1}^{-1}g_{2}\in S$ and so $g_{2}$ is a product of elements in $S$. Continuing in this manner we see that $g_{n}=g$ must also be a product of elements of $S$.
My question is how do we know that such a sequence $\{\mathcal{F}, g_{1}\cdot \mathcal{F}, g_{2}\cdot \mathcal{F}, \ldots, g_{n}\cdot \mathcal{F} = g\cdot \mathcal{F} \}$ satisfying the second condition must actually exist? For example why is it not possible that the intersection would necessarily be empty for any $g_{i+1}\neq g_{i}$?