Let's first spell out definitions and your question:
Suppose that $X$ is a metrizable topological space. For $G<Homeo(X)$ and $x\in X$, the subgroup $G_x=\{g\in G: gx=x\}$ is called the $G$-stabilizer of $x$.
A subgroup $G< Homeo(X)$ is said to be "properly discontinuous" (in Katok's sense) if for every compact $K\subset X$ and every $x\in X$ the subset
$$
\{g\in G: gx\in K\}\subset G
$$
is finite. This condition combines two properties: (a) finiteness of $G$-stabilizers of points in $X$, (b) the condition that every $G$-orbit in $X$ has no accumulation points in $X$. Moreover, (a) & (b) imply proper discontinuity in Katok's sense. (See my answer here.)
Your question: Suppose that $X$ is a metrizable space, $G<Homeo(X)$ is a subgroup such that for (i) all point-stabilizers $G_x<G$ are finite, and (ii) for every $x\in X$ the orbit $Gx\subset X$ is discrete (i.e. is a discrete space with respect to the subspace topology). Does it follow that $G$ acts properly discontinuously on $X$ in Katok's sense? Equivalently, does it follow that each $G$-orbit is closed in $X$?
I will show that the answer is negative. My example, in fact, will be coming from Fuchsian groups. Let $S$ be a orientable compact hyperbolic surface, $G< PSL(2, {\mathbb R})$ a discrete (Fuchsian) subgroup such that ${\mathbb H}^2/G$ is isometric to $S$. It is known that in this case every point of the boundary circle of the hyperbolic plane ${\mathbb H}^2$ is a point of approximation or, equivalently, conical limit point of the $G$-action on ${\mathbb H}^2$. See for instance Theorem 12.4.5 in
Ratcliffe, John G., Foundations of hyperbolic manifolds, Graduate Texts in Mathematics 149. Cham: Springer (ISBN 978-3-030-31596-2/hbk; 978-3-030-31597-9/ebook). xii, 800 p. (2019). ZBL1430.51002.
A better place to read this staff is in
Bowditch, B. H., Geometrical finiteness for hyperbolic groups, J. Funct. Anal. 113, No. 2, 245-317 (1993). ZBL0789.57007.
Here is the definition of "conical" (you can also find it in the above references):
Let $Z={\mathbb H}^2\sqcup S^1$ (where $S^1$ is the boundary circle of the hyperbolic plane). The $r$-conical topology $T_r$ (where $r>0$) on $Z$ is defined (via its basis) as follows: For points $z\in {\mathbb H}^2$ the topology is standard, the basis at $z$ consists of open disks $B(z,R)=\{x\in {\mathbb H}^2: d(x,z)<R\}$, where $R>0$ and $d$ denotes the hyperbolic metric. For points $z\in S^1$ the basis consists of the following subsets:
$$
C(z, \rho, r)=\{z\}\cup N_r(\rho),
$$
where $\rho\subset {\mathbb H}^2$ are hyperbolic rays asymptotic to $z$ and $N_r$ denotes the open $r$-neighborhood of $\rho$ in ${\mathbb H}^2$ with respect to the hyperbolic metric $d$ (the number $r$ is fixed as above). The terminology "conical" comes from the fact that if $\rho$ happens to be a Euclidean line segment $oz$ (in the unit disk model), connecting the center of the unit disk to $z$, then, apart from a bounded subset of ${\mathbb H}^2$, $N_r(\rho)$ is a Euclidean cone with the axis $oz$.
In any case, this basis generates a topology, denoted $T_r$, on $Z$. This is not the standard topology homeomorphic to the closed unit disk. In this topology, $S^1$ (with the subspace topology) is a discrete space, since for each $z\in S^1$, $C(z, \rho, r)\cap S^1=\{z\}$, i.e. $\{z\}$ is open in $S^1$.
Now, every group $G$ of hyperbolic isometries acts on $Z$ by homeomorphisms (where the action on the boundary circle is the standard one): In fact, $G$ preserves the above basis of topology on $Z$. Furthermore, $(Z, T_r)$ is Hausdorff for all $r>0$. The only mildly nontrivial case to consider is of $z_1\ne z_2\in S^1$. Take two rays $oz_i$ asymptotic to $z_i$ (still, the unit disk model) and remove from both some nonempty initial subsegments; the result are rays $\rho_i=x_i z_i$, $x_i\in oz_i$, $x_i\ne o$. Then
$$
C(z_1, \rho_1, r)\cap C(z_2, \rho_2, r)=\emptyset ,
$$
provided that $d(x_i,o)$ are large enough.
With a bit more work, one proves that for every $r>0$, $(Z,T_r)$ is a regular topological space, i.e. for each $z\in Z$ and a closed subset $F\subset Z\setminus \{z\}$, there are disjoint open neighborhoods $U(z), U(F)$ of $z, F$ in $(Z,T_r)$. The most interesting case is when $z\in S^1, F\subset S^1$. Let me know if you have trouble proving regularity.
Remark. Even though it is not needed, each space $(Z,T_r)$ is locally compact.
Let $z\in S^1$ be a point not fixed by any nontrivial element of $G$ (the set of fixed points is countable, hence, such $z$ always exists). Lastly, set
$$
X:= {\mathbb H}^2 \cup Gz \subset Z,
$$
equipped with the subspace topology induced from $T_r$ ($X$ is open in $Z$ with respect to $T_r$). The space $X$ is easily seen to be 2nd countable; it is also Hausdorff and regular as a subspace of Hausdorff and regular space $Z$. Hence, Urysohn's metrization theorem implies that $X$ is metrizable.
The point $z$ is an $r$-conical limit point of every $G$-orbit $Gx\subset {\mathbb H}^2$, meaning that there exists $r>0$ and a sequence $g_n\in G$ such that $g_nx\to z$ in the topology $T_r$. (This is the content of the theorem quoted above.) Thus, the orbit $Gx$ is discrete but not closed in $(X,T_r)$. For each $y\in Gz$, orbit $Gy$ is contained in $S^1$ and, hence, is discrete. Thus, every $G$-orbit in $(X,T_r)$ is discrete. At the same time, the $G$-stabilizer of every point in $X$ is trivial. Putting it all together, we obtain the required example: $G< Homeo(X, T_r)$ is a subgroup satisfying (i) and (ii) but not acting properly discontinuously in Katok's sense. What is it good for, I do not know.
