Is this a general way to show that a group action on a manifold is properly discontinuous without fixed points? How do I identify the quotient space? First, some definitions so that I am sure that it is clear what I am talking about:

Definition. Let $X$ be a manifold and $G$ a subgroup of $\operatorname{Diff}(X)$. We say that $G$ acts properly discontinuous without fixed points if the following conditions are satisfied:
(i) For every $x, y\in X$ such that $y$ is not in the orbit $G\cdot x$ of $x$ there exist neighborhoods $U\ni x$ and $V\ni y$ such that, for every $g\in G$, the intersection $g\cdot U\cap V$ is empty.
(ii) For every $x\in X$ there exists an open neighborhood $U \subset X$ of $x$ such that, for every $g\in G$ distinct from the identity, the intersection $g\cdot U\cap U$ is empty.

(I took this definition from page 68 of Berger's and Gostiaux's book Differential Geometry: Manifolds, Curves and Surfaces)
Now I would like to show that the action of $\mathbb{Z}^2$ on $\mathbb{R}^2$ defined for every $(m, n)\in \mathbb{Z}^2$ and $(x, y)\in \mathbb{R}^2$ by $$(m, n) \cdot (x, y)=(x+m+na, y+n),$$ where $a\ge 0$ is fixed, is free and properly discontinuous. The argument below is basically contained in the book I mentioned above (I say basically because they are considering the standard integral shift on $\mathbb{R}^N$, not this slightly modified version), but I have a couple of questions about it.
To check (i), take first $x, y\in \mathbb{R}^2$ such that $y\notin G\cdot x$. Define $\varepsilon=\inf\{d(y, z): z\in G\cdot x\}$, where $d$ denotes the standard Euclidean distance. Then, the book suggests that $\varepsilon>0$ "since there are only a finite number of points of $G\cdot x $ in any compact subset of $\mathbb{R}^2$". I don't really understand the part within the quotation marks. I agree that $\varepsilon>0$ because it is obviously nonnegative and, if it were $0$, then I think we would have that $G\cdot x$ is dense in the plane and this is clearly not the case if we try to visualize this orbit. Is my justification correct? Why does theirs work? Is considering the distance between $y$ and $G\cdot x$ the standard way to check this condition? To me it feels like it always does when our manifold is also a metric space (and in all the examples I encountered it was since all of them were on $\mathbb{R}^N$ or something embedded into $\mathbb{R}^N$).
To check (ii), if we fix some $x\in \mathbb{R}^2$, then we can just take $U=B\left(x, \frac{1}{2}\right)$ and we are done. We can actually take anything $<1$ instead of $\frac{1}{2}$ (or even any real number that is not a positive integer perhaps?) and this still works and it is geometrically clear to me why it does. I also believe that this is the general way to go when we act on a metric space (here it is quite convenient that our action is an isometry, so, once we see where $x$ goes, we basically have $g\cdot U$). Am I right?
Furthermore, I am asked to show that the quotient $\mathbb{R}^2/\mathbb{Z}^2$ constructed above is diffeomorphic to $T^2\subset \mathbb{R}^4$. I never understood how to identify these quotient manifolds. My instructor always did some drawings that I couldn't make much sense of and he said that he "glues" some things together, but I never really understood what he was doing.
P.S. I am aware of topics such as this one, where my hunch that we can generally proceed like above to prove that some action is properly discontinuous without fixed points, but I want to make sure that this is the case and I want to understand why the argument in the book for (i) works and whether mine based on density is also right.
 A: The goal of this answer is to address the second question about determining the diffeomorphism type of the quotient.  But first, a disclaimer:  I'm much more familiar with this question in the setting of determining the diffeomorphism type of the quotient $M/G$ when $G$ is a compact Lie group.
Here are a spattering of tools I'm familiar with:

*

*Find a smooth function $f:M\rightarrow X$ which is $G$-invariant.  We then automatically obtain a smooth function $g:M/G\rightarrow X$ for which $f = g\circ \pi$, where $\pi:M\rightarrow M/G$ is the projection.  In nice situations, we can prove things about $g$ using $f$.  For example, if $f$ is surjective, so is $g$.  If $f$ is injective on orbits (that is, $f(x) = f(y)$ implies $x = gy$ for some $g\in G$), then $g$ is injective. To ensure the inverse is smooth, one can often argue as follows:  show $f$ is a submersion.  This, then, implies $g$ is a submersion which often then implies $g$ is an injective local diffeomorphism, so is a diffeomorphism.

Off hand, the only example I know of which does this is a joint paper of  David González-Álvaro and myself.


*If $G$ has subgroups, or if the $G$-action on $M$ extends to a larger group action, then $M/G$ fits into a fiber bundle.  Specifically, if $H\subseteq G$, then we obtain the bundle $G/H\rightarrow M/H\rightarrow M/G$ (where $G/H$ is just a coset space since $H$ may not be normal).  If $G\subseteq K$ and the $G$-action on $M$ extends to a $K$-action on $M$, then $M/G$ is the total space of a bundle over $M/K$.  In both situations, one try to determine the diffeomorphism type of $M/H$ or $M/K$, and then use the bundle structure to analyze the diffeomorphism type of $M/G$.

A relatively famous example of this is due to Gromoll and Meyer.  They study an action of the Lie group $G = Sp(1)$ on $M=Sp(2)$, showing that the $G$ action extends to an action by $K = Sp(1)\times Sp(1)$.  Thus, they find the $M/G$ is the total space of an $S^3$-bundle over $M/K$.  Now, $M/K$ is a $4$-manifold, which are quite well understood - it's not too hard for them to show that $M/K$ is diffeomorphic to $S^4$.  Thus, $M/G$ is an $S^3$-bundle over $S^4$, which have been well studied (beginning with Milnor's paper where he establishes the existence of exotic spehres.)  In particular, Gromoll and Meyer were able to show that $M/G$ happens to be an exotic sphere.


*In dimension at most $5$, we have good classifications of low dimensional manifolds with sufficiently good control on the fundamental group (in particular, if the manifolds are simply connected).  In dimension $4$, this classification is up to homeomorphism and is due to Freedman. In dimension $5$, it's up to diffeomorphism and is done by Barden and Smale.  So, if you know $M/G$ is low enough dimensional, one can often compute things like the cohomology ring or characteristic classes to determine the diffeomorphism type.

Even in dimension $4$, this is often feasible, despite the gap between homeomorphisms and diffeomorphisms in that dimension.  This is a consequence of the fact that a $4$-dimensional quotient $M/G$ often has more structure than a generic $4$-manifold - maybe you know something about a bundle structure, or maybe you know something geometric.  It is often the case that while exotic versions of manifolds exist in dimension $4$, the exotic types can't carry the "nice" structure.  For example, while $\mathbb{R}^4$ admits three Riemannian metrics with sectional curvature respectively strictly positive, strictly negative, and zero, every Riemannian metric on an exotic $\mathbb{R}^4$ has some positive and some negative curvature.  For another example, while there could be exotic $S^2\times S^2$s, the only one with the structure of an $S^2$-bundle over $S^2$ is the standard $S^2\times S^2$.
Examples of this abound in papers dealing with low dimensional classifications.  One example would be the thesis of my student Andrew Lutz.


*Often for higher dimensional quotients, the answer is "$M/G$ is not diffeomorphic to any previously studied manifold."  In other words, often $M/G$ is our best description of that manifold.  This kind of thing happens all the time with homogeneous spaces.  For example, if we consider the action of $G=S^1$ on $M=SU(10)$ given by left multiplication by the diagonal matrix with entries $(z^2, z^4, z^{-5}, z^6, z^{-7}, 1, 1, 1, 1, 1)$ with $z\in S^1\subseteq \mathbb{C}$, then I'd be willing to bet that the space $M/G$ has never been studied before.  (I just made up the exponents, there's nothing particularly special about them.)

Edit.
For (i), if $G\cdot x$ were close to $y$ for every $y$,then you're right that $G\cdot x$ would be dense.  (By "close to" I mean "$\inf d(G\cdot x, y) = 0$").  But it's possible that $G\cdot x$ is close to some $y$ without being close to a all $y$s.  As a dumb example of this, if $G = \mathbb{Q}$ acting on $\mathbb{R}^2$ by $q(s,t) = (q+s,t)$, then the orbit $G\cdot (0,0)$ is close to all points along the $x$-axis, but far from any point not on the $x$-axis.
For the example under consideration, you're right that this $\inf$ must be positive.  Here's a more rigorous proof.  First, note that for any non-identity $g\in G$ and any $x\in \mathbb{R}^2$, that $d(g\cdot x,x) = \sqrt{(m+na)^2 + n^2 }\geq 1$.  Now, I claim that for any $y\notin G\cdot x$, the ball of radius $1/2$ around $y$ contains at most one point of $G\cdot x$.  For if both $g_1\cdot x$ and $g_2\cdot x$ are in the ball, then the triangle inequality gives $d(g_1\cdot x,g_2\cdot x)\leq d(g_1\cdot x,y)+d(y,g_2\cdot x) < 1/2 + 1/2 = 1$, contradicting the "First, note" sentence.  Now, if there are no points of $G\cdot x$ in the ball, then the $\inf$ is at least $1/2 > 0$, and if there is one point, then the $\inf$ is the distance between $y$ and that one point, so is also non-zero.
For ii), you're right.
