In the book An Introduction to Differentiable Manifolds and Riemannian Geometry by Boothby in Chapter $3$ the author gives the following definition:
Definition($8.1$) A discrete group $\Gamma$ is said to act properly discontinuously on a manifold $M$ if the action is $C^\infty$ and satisfies the following two conditions:
(i) Each $x\in M$ has a neighborhood $U$ such that the set $\{h\in \Gamma|hU\cap U\neq \emptyset\}$ is finite;
(ii) If $x, y \in M$ are not in the same orbit, then there are neighborhoods $U , V$ of $x , y$ such that $U\cap \Gamma V =\emptyset$.
In Exercise $3$ of this chapter the auther want to show that the following statement can be replaced by (i):
(i') The isotropy group $\Gamma_x$, of each $x\in M$ is finite, and each $x$ has a neighborhood $W$ such that $hW\cap W=\emptyset $ if $h\notin \Gamma_x $ , and $hW=W$ if $h\in \Gamma_x$.
Clearly (i') implies (i). I want to show that (i) implies (i') and I stuck here. I know that (i) shows that $\Gamma_x$ is finite and I could find a nbd of $x$ such that $hU=U$ for $h\in \Gamma_x$ (If $\Gamma_x=\{h_1,...,h_n\}$ then $W=\cap_i^n(U\cap h_iU)$ satisfies $hW=W$ if $h\in \Gamma_x$ ) but I have o problem to show that $hU\cap U=\emptyset $ if $h\notin \Gamma_x $.
You will also need other portions of the definition (in my previous draft I said you would need (ii), but now I think you only need Hausdorff). To say that (i) can be replaced by (i') in that definition does not necessarily mean (i) $\iff$ (i'). It means instead that the definition with (i) is equivalent to the definition with (i').
Applying (i), choose a neighborhood $U$ of $x$ so that the subset $\{h \in \Gamma \, \bigm| \, hU \cap U \ne \emptyset\}$ is finite. Some elements of that subset are in $\Gamma_x$ and some are in $\Gamma-\Gamma_x$; let's enumerate the elements of that subset that are in $\Gamma-\Gamma_x$ as $h_1,...,h_K$. Denote points $y_1=h_1 x$, ..., $y_K=h_K x$, all of which are different from $x$. Using that a manifold is Hausdorff, for each $k=1,...,K$ choose disjoint neighborhoods $U_k$ of $x$ and $V_k$ of $y_k$. Let $$W' = U \cap U_1 \cap \cdots \cap U_K \cap h_1^{-1} V_1 \cap \cdots \cap h_K^{-1} V_K $$ Note that for all $h \in \Gamma-\Gamma_x$ we have $hW' \cap W' = \emptyset$:
Finally, take $W = \bigcap_{h \in \Gamma_x} h W'$.