I know of several different arguments. You can decide which one you think is most elegant...

1. Rohlin's argument, which is actually quite geometric. You start with an immersion of the 3-manifold in $\mathbb{R}^5$. You modify the immersion by a cobordism until it is an embedding, and then find an explicit 4-manifold bounding it. This is nicely explained in "A la recherche de la topologie perdue". I believe this is also Autumn Kent's answer above.

2. Thom's argument, with lots of algebraic topology. This is probably not the most elegant route if you only want this piece, although of course Thom tells you much more.

3. Rourke's argument as sketched by Daniel Moskovich above. Indeed, any proof that the mapping class group is generated by Dehn twists also gives a proof that $\Omega_3 = 0$. Dehn and Lickorish also have proofs of this.

4. I also have a proof with Francesco Costantino, also direct and geometric. You take the compact 3-manifold and look at a generic map to $\mathbb{R}^2$. The preimage of a generic point is a disjoint union of circles, which bounds a convenient canonical surface (a union of disks). Take these disks as the start of your 4-manifold. In codimension one singularities, two of these circles can merge, and the preimage of a little transversal is a pair of pants, which can be filled in with a 3-sphere (together with the disks already attached). In codimension 2, there are only two different interesting local models, and both can be filled in canonically with a 4-ball.

The reason to prefer our proof (number 4) is that it is more efficient, in that (e.g.) for a 3-manifold triangulated with $n$ tetrahedra, it gives a 4-manifold with bounded geometry with $O(n^2)$ simplices. By comparison, the mapping-class group arguments of (3) tend to give a 4-manifold of complexity at least exponential in $n$, and usually a tower of exponentials. (You can see this already in the inductive argument sketched out in Daniel Moskovich's answer.) Thom's proof (2) is completely non-explicit; I don't know how to extract any bounds from it. Rohlin's proof (1) can, I believe, be shown to give a 4-manifold with $O(n^4)$ simplices, although I never worked out all the details.

I know of several different arguments. You can decide which one you think is most elegant...

1. Rohlin's argument, which is actually quite geometric. You start with an immersion of the 3-manifold in $\mathbb{R}^5$. You modify the immersion by a cobordism until it is an embedding, and then find an explicit 4-manifold bounding it. This is nicely explained in "A la recherche de la topologie perdue". I believe this is also Autumn Kent's answer above.

2. Thom's argument, with lots of algebraic topology. This is probably not the most elegant route if you only want this piece, although of course Thom tells you much more.

3. Rourke's argument as sketched by Daniel Moskovich above. Indeed, any proof that the mapping class group is generated by Dehn twists also gives a proof that $\Omega_3 = 0$. Dehn and Lickorish also have proofs of this.

4. I also have a proof with Francesco Costantino, also direct and geometric. You take the compact 3-manifold and look at a generic map to $\mathbb{R}^2$. The preimage of a generic point is a disjoint union of circles, which bounds a convenient canonical surface (a union of disks). Take these disks as the start of your 4-manifold. In codimension one singularities, two of these circles can merge, and the preimage of a little transversal is a pair of pants, which can be filled in with a 3-sphere (together with the disks already attached). In codimension 2, there are only two different interesting local models, and both can be filled in canonically with a 4-ball.

The reason to prefer our proof (number 4) is that it is more efficient, in that (e.g.) for a 3-manifold triangulated with $n$ tetrahedra, it gives a 4-manifold with bounded geometry with $O(n^2)$ simplices. By comparison, the mapping-class group arguments of (3) tend to give a 4-manifold of complexity at least exponential in $n$, and usually a tower of exponentials. (You can see this already in the inductive argument sketched out in Daniel Moskovich's answer.) Thom's proof (2) is completely non-explicit; I don't know how to extract any bounds from it. Rohlin's proof (1) can, I believe, be shown to give a 4-manifold with $O(n^4)$ simplices, although I never worked out all the details.

I know of several different arguments. You can decide which one you think is most elegant...

1. Rohlin's argument, which is actually quite geometric. You start with an immersion of the 3-manifold in $\mathbb{R}^5$. You modify the immersion by a cobordism until it is an embedding, and then find an explicit 4-manifold bounding it. This is nicely explained in "A la recherche de la topologie perdue". I believe this is also Richard Kent's answer above.

2. Thom's argument, with lots of algebraic topology. This is probably not the most elegant route if you only want this piece, although of course Thom tells you much more.

3. Rourke's argument as sketched by Daniel Moskovich above. Indeed, any proof that the mapping class group is generated by Dehn twists also gives a proof that $\Omega_3 = 0$. Dehn and Lickorish also have proofs of this.

4. I also have a proof with Francesco Costantino, also direct and geometric. You take the compact 3-manifold and look at a generic map to $\mathbb{R}^2$. The preimage of a generic point is a disjoint union of circles, which bounds a convenient canonical surface (a union of disks). Take these disks as the start of your 4-manifold. In codimension one singularities, two of these circles can merge, and the preimage of a little transversal is a pair of pants, which can be filled in with a 3-sphere (together with the disks already attached). In codimension 2, there are only two different interesting local models, and both can be filled in canonically with a 4-ball.

The reason to prefer our proof (number 4) is that it is more efficient, in that (e.g.) for a 3-manifold triangulated with $n$ tetrahedra, it gives a 4-manifold with bounded geometry with $O(n^2)$ simplices. By comparison, the mapping-class group arguments of (3) tend to give a 4-manifold of complexity at least exponential in $n$, and usually a tower of exponentials. (You can see this already in the inductive argument sketched out in Daniel Moskovich's answer.) Thom's proof (2) is completely non-explicit; I don't know how to extract any bounds from it. Rohlin's proof (1) can, I believe, be shown to give a 4-manifold with $O(n^4)$ simplices, although I never worked out all the details.

I know of several different arguments. You can decide which one you think is most elegant...

1. Rohlin's argument, which is actually quite geometric. You start with an immersion of the 3-manifold in $\mathbb{R}^5$. You modify the immersion by a cobordism until it is an embedding, and then find an explicit 4-manifold bounding it. This is nicely explained in "A la recherche de la topologie perdue". I believe this is also Autumn Kent's answer above.

2. Thom's argument, with lots of algebraic topology. This is probably not the most elegant route if you only want this piece, although of course Thom tells you much more.

3. Rourke's argument as sketched by Daniel Moskovich above. Indeed, any proof that the mapping class group is generated by Dehn twists also gives a proof that $\Omega_3 = 0$. Dehn and Lickorish also have proofs of this.

4. I also have a proof with Francesco Costantino, also direct and geometric. You take the compact 3-manifold and look at a generic map to $\mathbb{R}^2$. The preimage of a generic point is a disjoint union of circles, which bounds a convenient canonical surface (a union of disks). Take these disks as the start of your 4-manifold. In codimension one singularities, two of these circles can merge, and the preimage of a little transversal is a pair of pants, which can be filled in with a 3-sphere (together with the disks already attached). In codimension 2, there are only two different interesting local models, and both can be filled in canonically with a 4-ball.

The reason to prefer our proof (number 4) is that it is more efficient, in that (e.g.) for a 3-manifold triangulated with $n$ tetrahedra, it gives a 4-manifold with bounded geometry with $O(n^2)$ simplices. By comparison, the mapping-class group arguments of (3) tend to give a 4-manifold of complexity at least exponential in $n$, and usually a tower of exponentials. (You can see this already in the inductive argument sketched out in Daniel Moskovich's answer.) Thom's proof (2) is completely non-explicit; I don't know how to extract any bounds from it. Rohlin's proof (1) can, I believe, be shown to give a 4-manifold with $O(n^4)$ simplices, although I never worked out all the details.