Katok's Fuchsian groups: Properly discontinuous action - homeomorphisms vs isometries hypotheses

I want to understand Katok's Fuchsian groups (page 28) proof of the theorem:

Theorem 2.2.1: $G$ acts properly discontinuously on $X$ if and only if each point $x\in X$ has a neighborhood $V$ such that $T(V)\cap V\neq \emptyset$ for only finitely many $T\in G$.

To her, $X$ is a metric space and $G$ is a group of homeomorphisms of $X$, not necessarily isometries. Right before this theorem, she comments that $G$ acts properly discontinuously on $X$ if, and only if, each $G$-orbit, namely $G(x)=\{g(x)\in X\,|\,g\in G\}$, is a discrete subset of $X$ and the stabilizer, namely $G_x=\{g\in G\,|\,g(x)=x\}$, of each point $x\in X$ is finite. This fact is okay, I could prove it, just considering $G$ as a group of homeomorphisms.

Let's take a look at Katok's proof of theorem 2.2.1:

Proof: Suppose $G$ acts properly discontinuously on $X$, then each $G(x)$ is discrete, and for each point $x\in X$, $G_x$ is finite. This implies that for any point $x$ there exists a ball $B_\epsilon(x)$ centered at $x$ of radius $\epsilon$ containing no points of $G(x)$ other than $x$. Let $V\subset B_{\epsilon/2}(x)$ be a neighborhood of $x$, $\underline{\textbf{then $T(V)\cap V\neq \emptyset$ implies that $T\in G_x$}}$, hence it's possible for only finitely many $T\in G$. [...]

I couldn't see why $T$ should be in $G_x$. In fact, making some drawings I started to doubt of the validity of this claim (since $T$ is just a homeomorphism, it can distort metric and "melt" all the figure, circles, etc). Although, if $G$ is a group of isometries, the proof of this fact is easy...

So this is my question: is it suffice $G$ to be a group of homeomorphisms in order to ensure Theorem 2.2.1 or is it needed to require it is a group of isometries instead?

I've saw some similar questions on the site, but I want to know if the homeomorphism hypothesis is or not sufficient to get the result... This is not clear to me yet.

Edit: If, in addition to the homeomorphism hypothesis, $G$ is also equicontinuous, then $G$ acts properly descontinuously if, and only if, there is such a neighborhood $V$. But is necessary to require this? What is the minimal set of hypothesis that I need to add in order to have the theorem?

Edit 2: A group of isometries is, of course, (uniformly) equicontinuous.

Edit 3: Definition:(Katok's definition of properly discontinuous action) We say that a group $G$ acts properly discontinuously on $X$ if the $G$-orbit of any point $x\in X$ is locally finite.

This means that, for any compact set $K\subset X$ the set $\{g\in G\,|\,g(x)\in K\neq \emptyset\}$ is finite, and this holds for every $x\in X$.

I will be assuming that $X$ satisfies the T1 condition (every point is closed). If you are interested in more general spaces, I would like to hear an explanation why.

Consider the following three conditions:

D1: Katok's proper discontinuity condition: Point-stabilizers are finite and for every compact $K\subset X$, every $x\in X$, the set $\{g\in G: gx\in K\}$ is finite.

D2. Point-stabilizers are finite and every $G$-orbit in $X$ is discrete and closed, i.e. has no accumulation points.

D3. For each $y\in X$ there is a neighborhood $U=U_y$ such that $\{g\in G: gU\cap U\ne\emptyset\}$ is finite.

Proposition 1. D3$\Rightarrow$D2. (This implication is also proven in this question, where it is also explained why the T1 condition is needed.)

Proof. Suppose that some $G$-orbit $Gx\subset X$ accumulates at a point $y\in X$: Since $X$ is T1, there are infinitely many elements $g_i\in G$ such that $x_i=g_ix\in U$ for every neighborhood $U$ of $y$. Let $U=U_y$ be a neighborhood of $y$ as in D3. Then, by looking at the elements of the form $h_{ij}=g_i g_j^{-1}\in G$ we see that $g_{ij}U\cap U\ne\emptyset$ ($h_{ij}$ maps $x_j$ to $x_i$). A contradiction. qed

Proposition 2. D2$\Rightarrow$D1.

Proof. Consider a compact $K\subset X, x\in X$ and the subset $K_x:= K\cap Gx$. Then $K_x$ is a discrete closed subspace of a compact space. But every discrete closed subspace of a compact space is finite. Hence, D1 follows. qed

Proposition 3. If you assume, in addition, that $X$ is locally compact, or 1st countable, or that $G$ is countable, then D1$\Rightarrow$D2.

Proof. I will prove it assuming that $X$ is 1st countable only. Suppose that $Gx$ accumulates at a point $y\in X$. Then, since $X$ is 1st countable, there is an infinite sequence $g_n\in G$, such that $g_nx$ converges to some $y\in X$. The subset $K:= \{y\}\cup \{g_nx: n\in {\mathbb N}\}$ is compact and $g_nx\in K$ for all $n$. This contradicts D1. qed

Proposition 4. D2 does not imply D3 even for cyclic groups of homeomorphisms of surfaces.

Proof. Consider the "standard example" (as in the link I gave earlier). Namely, let $g: (x,y)\mapsto (2x, y/2)$ be the linear map of the real plane as in that example; let $G$ be the cyclic group generated by $g$. Restrict the $G$-action to the subset $Q\subset {\mathbb R}^2$, which is the first coordinate quadrant $x\ge 0, y\ge 0$ with the origin removed. Clearly, all the orbits of the $G$-action on $Q$ are discrete and point-stabilizers are trivial. Now, form the quotient of $Q$ by the equivalence relation $$ (x,0)\sim (0, 1/y). $$ This quotient $A$ is homeomorphic to the punctured plane. The map $(x,0)\mapsto (0, 1/y)$ is $G$-equivariant, hence, the $G$-action on $Q$ descends to a $G$-action on $A$. Again, it is clear that point-stabilizers for the $G$-action on $A$ are trivial and orbits are discrete. However, for each point $p\in A$ which is the projection of some $(x,0)\in Q$, and each neighborhood $U$ of $p$, the subset $\{g\in G: gU\cap U\ne \emptyset\}$ is infinite (for the same reason that the $G$-action on $Q$ is not proper in the conventional sense). qed

Lastly, I really dislike Katok's definition of proper discontinuity. It is not strong enough to guarantee that $X/G$ is Hausdorff (when $X$ is Hausdorff); I suspect it is not even strong enough to imply that $X/\Gamma$ is T1 (without extra assumptions as in Proposition 3).