Let $g$ be some homogeneous Riemannian metric on $\mathbb R^3$ (i.e, a metric such that $(\mathbb R^3,g)$ is a homogeneous space). Consider its isometry group $G = Isom(\mathbb R^3,g)$.
A covering subgroup $H \subset G$ is a group acting freely and properly discontiously on $\mathbb R^3$, in other words, the natural projection $\pi: \mathbb R^3 \to \mathbb R^3/H$ is a covering map. Note that trivial group $H = \{e\}$ is always a covering subgroup of $G$.
Question 1: Does there exist a cocompact covering group $H \subseteq G$, i.e a covering group such that $\mathbb R^3/H$ is compact ?
Suppose $G$ contains a cocompact covering group. Then it is a classic result by Thurston that, up to conjugation by some diffeomorphism $\phi: \mathbb R^3 \to \mathbb R^3$, there are six Riemannian metrics $\{g_i\}_{i=1}^6$ on $\mathbb R^3$ (independent of the choice of $g$), so that $G$ is contained in precisely one of $Isom(\mathbb R^3,g_i)$.
Question 2: Is there some $G$ as above, such that (a conjugate of) $G$ is properly contained in one $Isom(\mathbb R^3,g_i)$, i.e $G \subsetneq Isom(\mathbb R^3,g_i)$.
If we consider $\mathbb R^2$ instead of $\mathbb R^3$, such a homogeneous metric cannot be found. For if $g$ is any homogeneous metric on $\mathbb R^2$, then as $\mathbb R^2$ is two dimesional, this means that $(\mathbb R^2,g)$ is also an isotropic space, i.e, $G$ acts transitively on the set of all tangent planes in $T\mathbb R^2$. It follows that $(\mathbb R^2,g)$ has constant curvature, implying that $G$ is either conjugate to $Isom(\mathbb R^2,g_{eukl})$ or $Isom(\mathbb R^2,g_{hyp})$, both of which are maximal.
EDIT: In an earlier version, I mistakenly assumed that every metric $g$ on $\mathbb R^3$ would induce a homogeneous structure, simply because every translation is always an isometry, regardless of the specific choice of $g$. This is, of course, blatantly false, as Professor Lee was so nice to remind me of, it's because the inner product induced by $g$ may very well vary from point to point (as it's the usual case when picking a random metric on a manifold).
Question 1: No. For instance, take the Riemannian metric on $R^3$ $$ dt^2 + e^{at}dx^2 + e^{bt}dy^2 $$ for generic values of $a, b>0$. This geometry is homogeneous (with a 3-dimensional solvable Lie group acting transitively) but has no discrete cocompact subgroups of the isometry group.
Question 2: Yes. For instance, start with the 3-dimensional real Heisenberg group $H_3$ and equip it with a suitable left-invariant Riemannian metric which is also invariant under $SO(2)$ (this is known as the Nil-geometry, see Thurston's book that you are reading). Then $H_3$ contains a discrete cocompact subgroup, while $H_3$ is strictly smaller than $Isom(H_3)$.