Does every simply connected, orientable, non-compact, 3-manifold embed in $\mathbb{R}^3$? Let $M$ be a simply connected, (orientable), non-compact, 3-manifold without boundary. Must $M$ be homeomorphic with a topological subspace of $\mathbb{R}^3$?
 A: When the manifold is the universal cover of a compact $3$-manifold $M$ (to begin with, lets say without boundary) then you construct the embedding by hands, using geometrization.  In your question let's replace $\Bbb R^3$ with $S^3$ and we will see the embedding into $\Bbb R^3$ comes for free in the situation you are interested in.
An instructive case comes from the  case where your manifold is a connect-sum of lens spaces.   This has been the subject of a few recent threads:
Universal covers of non-prime 3-manifolds
The basic idea is that you choose a collection of reducing spheres for the connect sum decomposition, call them $\Sigma$.  Then $M \setminus \Sigma$ is a disjoint union of punctured lens spaces.  Each of these have universal covers diffeomorphic to  punctured spheres, so they embed in $S^3$.  You then glue the embedded punctured spheres together (in $S^3$) so that the appropriate boundary spheres are glued together.  If you view the embedded punctured spheres from the perspective of their complements in $S^3$ we are essentially doing the canonical construction of a Cantor set.  There  is the exceptional case of $\Bbb RP^3 \# \Bbb RP^3$ where we are constucting the standard embedding $S^0 \to S^3$.
But this is the basic idea.  The remaining geometric 3-manifolds have universal covers that are also subsets of $S^3$, so you similarly glue these together along  the (lifted) torus decomposition or sphere decompositions.  The tori (being incompressible) will lift to copies of $\Bbb R^2$, so those gluings are a little easier to visualize.
