Do all closed (not necessarily orientable) 3-manifolds embed into some closed orientable 4-manifold? Let $M^3$ be a closed not necessarily orientable 3-manifold.  Does $M$ embed (smoothly - though hopefully this does not matter) into some smooth closed orientable 4-manifold $X^4$?  
If $M$ is orientable, then of course since $\Omega_3 =0$, we can find a orientable 4-manifold that bounds $M$ and taking the double of this 3-manifold gives the desired $X$ - but what do I do when $M$ is not orientable?
 A: If $M$ is nonorientable, I believe the following modification of Cheerful Parsnip's idea works: 
By nonorientability of $M$, the first Stiefel-Whitney class $w_1(TM) \neq 0 \in H^1(M; \Bbb Z_2)$. Since $H^1(M; \Bbb Z_2) = \text{Hom}(\pi_1 M, \Bbb Z_2)$, this can be interpreted as a homomorphism $w_1 : \pi_1 M \to \Bbb Z_2$ which inputs a (homotopy class of) loop $\gamma$ based at some point $p$ in $M$ and outputs $+1$ or $-1$ depending on if that preserves or reverses a chosen orientation at $p$ once transported along the loop. $w_1$ gives rise to a real line bundle on $M$, so let's denote that as $(E, M, \pi)$. Note that this is the determinant bundle of the tangent bundle of $M$.
There's a (non-canoical) decomosition of the tangent bundle of the total space as $TE \cong \pi^* E \oplus \pi^* TM$, so we have equalities $$w_1(TE) = w_1(\pi^* E) + w_1(\pi^* TM) = \pi^* (w_1(E) + w_1(TM))$$ But since $E = \text{det}(TM)$, $w_1(E) = w_1(TM)$, which implies $w_1(E) + w_1(TM) = 0$ as an element of $H^1(M; \Bbb Z_2)$. Therefore, $w_1(TE) = 0$, which implies the total space $E$ is an orientable manifold. Roughly speaking, given a loop $\gamma$ in $E$, transporting an orientation at the basepoint around is either preserved along both the vertical and horizontal subbundle of $E$, or reversed along both, which implies it's preserved globally in $E$ because $1\cdot 1 = (-1)\cdot(-1) = 1$.
Equip $E$ with a bundle metric and let $E_{\leq 1} = \{v \in E : \|v\| \leq 1\}$ be the unit ball bundle. The boundary $\partial E_{\leq 1}$ is the unit sphere bundle, which is the orientation double cover $\tilde{M}$ of $M$. Take two copies of $E_{\leq 1}$ and glue them along the boundary by the identity diffeomorphism of $\tilde{M}$ to obtain an orientable closed manifold $X$ in which $M$ embeds via the zero section $M \to E$. Note the similarity with the orientable case, as $X$ is also a $S^1$-bundle over $M$.
