Suppose $G$ is a topological group, $X$ a topological space and $G \times X \rightarrow X$ group action that is continuous. Further, suppose that the action is free ($G_x = \{e\}$, for all $x$).
What I want to prove is the following: $$\forall x \in X. \exists U_x. \forall g, g' \in G. g \neq g' \Rightarrow g\cdot U_x \cap g' \cdot U_x = \emptyset$$
But I'm not entirely sure, if this is even true. (The converse however, is)
This statment is equivalent to the statement $\forall x \in X. \exists U_x. \forall g \in G. g \neq e \Rightarrow g\cdot U_x \cap \cdot U_x = \emptyset$, by translating your definition by $g^{-1}$.
Let $X = \mathbb{R}$, and $G = (\mathbb R, +)$ be the additive group acting on $X$ by translation. This is clearly a free action. Now for $x= 0\in X$, you would need an interval $0\in (-\epsilon, \epsilon)$ such that $\frac{1}{n}\notin (-\epsilon, \epsilon)$ for all $n$, since $\frac{1}{n}=\frac{1}{n}+0 \in \frac{1}{n}+(-\epsilon, \epsilon)$, which is impossible.
Thus, your statement does not hold.