Classifying point stabilizers for the groups associated with 3D model geometries. For those who have the book, this question is regarding p181 in Thurston's "Three Dimensional Geometry and Topology" (although I will do my best to summarize it).  Basically, there's an entire paragraph where all of the implications make no sense to me.
We have a connected, simply connected manifold $X$.  $G$ is a Lie group that acts transitively by diffeomorphisms on $X$, and has compact point stabilizers.  We also require that $G$ is maximal among all Lie groups satisfying these conditions.  The goal is to enumerate the $(G,X)$ pairs that serve as geometries for a 3-manifold, in the sense that there is a manifold with charts in $X$ such that the compatibility maps are local restrictions of elements of $G$.
The problem paragraph is as follows:

In... we will first look at the connected component of the identity of $G$ - call it $G'$.  The action of $G'$ is still transitive, and the stabilizers $G_x'$ of points $x \in X$ are connected.  This is because the quotients $G_x' / (G_x')_0$, where $(G_x')_0$ is the component of the identity of $G_x'$, form a covering space of $X$.  Since $X$ is simply connected, the covering is trivial.
Therefore $G_x'$ is a connected closed subgroup of $SO(3)$.

I assume the main purpose of restricting to the component of the identity is to cut out orientation reversing components (as in the case of $G = O(3)$). However, I have no idea what he means by the $G_x' / (G_x')_0$ being a covering of $X$.  In the example of $X = \mathbb{R}^3$ and $G$ the group of Euclidean isometries of 3 space, wouldn't we have $G'_x = (G'_x)_0 = SO(3)$, so the quotient would be trivial?  So this would be saying that trivial points are covering spaces of $\mathbb{R}^3$?  Even in cases where it wasn't trivial, how does it make sense for quotients of pieces of $G$ to be covering spaces for $X$ in a canonical way?
I'd greatly appreciate the help of anybody who can translate the paragraph.  Not really sure what tags to use here, since it's so specific.
 A: Here is my reading of this text:
Consider the topological space
$$
\tilde X= \bigsqcup_{x\in X} G'_x/(G'_x)_0. 
$$
This space has the natural topology coming from the fact that each $G'_x$ is a subset (in fact, a subgroup) of a topological group $G$. 
Edit: The natural topology I had in mind here is coming from the weakest topology on $\sqcup_{x\in X} G'_x$ in which the map from this space to $G'$ is continuous. Then pass to the quotient topology on $\tilde X$. But it is then unclear why this topology on $\tilde X$ is Hausdorff. See the edit below for a clean argument. 
There is a natural map 
$$
p: \tilde X\to X
$$ 
sending each $G'_x/(G'_x)_0$ to $x$. It is clear that this map is continuous; one then has to check that this map is also a covering map. One also needs to check that $\tilde X$ is connected (this follows from connectedness of $G'$). Since $X$ is simply-connected it then follows that $p$ is a homeomorphism. 
Edit: Here is a cleaner and complete argument. Pick a base-point $z\in X$. Then the natural map 
$$
G'/G'_z\to X
$$
induced by the orbit map $g\to gz$, is a diffeomorphism. Since $G'$ acts transitively on $X$, all subgroups $(G'_x)_0$ are conjugate to $(G'_z)_0$;  moreover, $K=(G'_z)_0$ is a normal subgroup in $G'_z$ with finite quotient group $F=G'_z/(G'_z)_0$. Now, consider the space $\tilde X= G'/(G'_z)_0$. The group $F$ acts on this quotient via right multiplication, for $f\in F$ define
$$
f\cdot gK =gKf= gfK.  
$$
We obtain diffeomorphisms 
$$
\tilde X/F\cong G'/G'_z\cong X. 
$$ 
One then sees that the action of the finite group $F$ is free, hence, the quotient map $p: \tilde X\to \tilde X/F$ is a covering map. The space $\tilde X$ is connected since $G'$ is. Therefore, $p$ is a covering map. Since $X$ is simply-connected, $p$ is a diffeomorphism. Therefore, $F=1$ and $G'_z$ is connected. 
Edit 2. In fact, one can eliminate the entire covering argument, by appealing to the long exact sequence of homotopy groups/sets of the fibration
$$
G_x\to G\to X
$$
which immediately shows that simple connectivity of $X$ and connectivity of $G$ imply connectivity of $G_x$. 
