Compactification theorem for differentiable manifolds ? Just parallelling this question, that seemed not to admit an easy answer at all, let's "soft down" the category and ask the same thing in the case of $\mathcal{C}^{\infty}$-differentiable manifolds [Edit: we consider only manifolds without boundary].
Well, so:

Is every differentiable manifold diffeomorphic to an open submanifold of a compact one?

Edit: As some comments have pointed out, there are manifolds for which the compactification theorem fails, so someone has suggested to change the question to the more meaningful:

Which differentiable manifolds are diffeomorphic to an open submanifold of a compact one?

 A: I think none of the above posts answer the question "is every differentiable manifold diffeomorphic to an open submanifold of a compact one?". Rather they answer "is every differentiable manifold diffeomorphic to the interior of a compact one?" The reason for the confusion could be the latter question is fundamental in geometric topology, while the former one has little significance. Anyway, 
The connected sum $V$ of infinitely many copies of $CP^3$'s is not diffeomorphic to an open subset of a compact manifold.
EDIT:  Hats off to Torsten Ekedahl who pointed out in comments that my argument below is incorrect (thus I don't know whether the above statement about $V$ is true). I decided not to delete it because it illuminates some subtleties of the original question.
The point is that any diffeomorphism onto an open subset pulls back the tangent bundle, and in  particular, pulls back the first Pontryagin class $p_1$. Thus if $V$ is an open subset of a compact manifold $M$, then its first Pontryagin class $p_1(V)$ lies in the image of $H^4(M)\to H^4(V)$, which is a finitely generated subgroup of $H^4(V)$, which is the infinite product of $\mathbb Z$'s corresponding to generators of $H^4(CP^3)$. The first Pontryagin class of $CP^3$ is a multiple of a generator of $H^4(CP^3)\cong\mathbb Z$, and removing a finite set of points from $CP^3$ does not affect the $4$th skeleton, so $p_1(V)$ does not lie in a finitely generated subgroup of $H^4(V)$.
I am curious to see low-dimensional answers to the question "is every differentiable manifold diffeomorphic to an open submanifold of a compact one?"
A: I'm trying to answer the second question. Most of the results can be found in the joint paper with Guilbault https://arxiv.org/abs/1712.05995.
Here I just wanted to make a quick summary.
An $m$-manifold $M^m$ with (possibly empty) boundary is completable if there exists a compact manifold $\widehat{M^m}$ and a compactum $C \subseteq \partial \widehat{M^m}$ such that $\widehat{M^m}-C$ is homeomorphic to $M^m$. In this case $\widehat{M^m}$ is called a (manifold) completion of $M^m$. One can change the homeomorphism to diffeomorphism or PL homeo. for other categories.
Dim = 0,1 are obvious.
Dim = 2, we can't find a complete classification in the literature, so we provided a theorem in that paper. That is, a connected 2-manifold is completable iff it has finitely generated first homology. This is mainly based on classical work of Kerekjarto and Richards.
Dim = 3, it's mainly due to Tucker, where he showed that a
3-manifold can be completed if and only if each component of each clean neighborhood
of infinity has finitely generated fundamental group.
Let me talk about dimensions $\geq 6$ first. Then I'll go back to dimensions 4 and 5. The first breakthrough regarding this problem was due to Siebenmann in 1965. In his PhD thesis, he proved that an open n-manifold $M^n$ is completable (it was called collarable by that time) iff
(1) M is inward tame, (2) the end is pro-$\pi_1$ stable and (3) the Wall finiteness obstruction of the end vanishes.
In 1983, O'Brien generalized the theorem to one-ended manifold with possibly non-empty boundary.
In our paper, we dropped the O'Brien's assumption on that manifolds are one-ended. By properly generalizing Siebenmann's conditions, we proved that manifolds of dimension at least 6 are completable iff they are inward tame, peripherally $\pi_1$-stable at infinity, of zero Wall and Whitehead torsion. Our proof is based on PL manifolds, but one can employ standard techniques such as "rounding off corners" to handle the other catergories.
Our theorem is still true in dimension = 5 provided that the fundamental groups are good in the sense of Freedman and Quinn.
The theorem fails in dimension = 4. Kwasik-Schultz and (independently) Weinberger discovered that there are open 4-manifold satisfying Siebenmann's condition but fail to be collarable.
Just a quick comment on Ian's reference to a contractible open manifold $M'$ constructed by Kister-McMillan which doesn't embed in $S^3$. The example was first proposed by R. H. Bing. Haken further proved that $M'$ doesn't embed in any 3-manifold using his finiteness theorem. Recently, I showed that $M'$ can't be embedded in any compact, locally connected and locally 1-connected metric space. https://arxiv.org/abs/1809.02628
A: No.  A surface of infinite genus is not a submanifold of a compact surface.
A: There are contractible 3-manifolds which cannot be embedded in any compact 3-manifold. Kister and McMillan constructed a variant of the Whitehead manifold $M'$ which is contractible but which cannot embed into $S^3$. From the Geometrization theorem, the universal cover of any compact 3-manifold embeds into $S^3$.  So if $M'$ embedded into a compact 3-manifold $M'\subset M$, its lift $M'\subset \widetilde{M}\subset S^3$ to the universal cover would give a contradiction. 
A: As everyone has said, the answer is "no".  You have to make assumptions to ensure that the "ends" of your manifold are sufficiently simple.  It appears hard to find using the refs Paul posted, but the key result about this is Larry Siebenmann's thesis.  I don't think this was ever published, but it is available on Andrew Ranicki's webpage here.  Another source (also on Ranicki's webpage) for this is some lecture notes of Kervaire, available here.
By the way, one obvious necessary condition is for your manifold to have a finitely presentable fundamental group (this is one of the problems with Richard's example).  A classic example to show that this is still not enough (even in dimension 3) is the Whitehead manifold.
EDIT : I should also point out one beautiful recent about this.  Marden's Tameness Conjecture (recently proved independently by Agol and Calegari-Gabai) says that if M is a hyperbolic 3-manifold with finitely generated fundamental group, then M is homeomorphic to the interior of a compact 3-manifold.  The Whitehead manifold mentioned above shows that the assumption that M is hyperbolic is necessary.
A: There is a long history on this problem, starting, in  dimensions>4 with 
Browder-Levine-Livesay:
http://www.jstor.org/stable/2373259?origin=crossref
http://www.ams.org/mathscinet-getitem?mr=189046 
Follow MR to get to results in dimensions 3,4, etc. You have to first eliminate 
issues like Richard mentions using finiteness obstructions.
