Does every torus $T\subset S^3$ bounds a solid tours $S^1\times D^2\subset S^3$? I want to show this by using Alexander's Theorem's proof method.
So here's what I thought. 
As I surger $T$, I have 2 $S^2$. So one bounds $S^1$ and the other bounds $D^2$. 
By reversing the surgery, I have that $T$ bounds a solid torus $S^1\times D^2$ in $S^3$.
Is it correct?
 A: "There's also a proof of this that involves basically following along with what Hatcher did for spheres and... doing it for tori. I don't remember the details, but that's how I did this argument when I looked at the book first."
Let $T$ be an embedded torus in $\mathbb{R}^3$. Then isotope $T$ so that $h:T\to \mathbb{R}$, the height function, is a Morse function. Nothing fancy here, push $T$ around so that none of the saddle points, maximums, and minimums are at the same height. Then slice $T$ along horizontal planes at heights which trap each critical value of $h$ between two planes. Since $T$ is closed and $2$ dimensional, the intersections of $T$ with the horizontal planes will be circles. By cutting $T$ along innermost circles in the intersection with the horizontal planes, pushing the circles off of the planes, and capping the wounds with disks, we surger $T$ into a bunch of components which are isotopic to spheres.
If all of our surgery loops, pre-surgery, bound disks in $T$, then it turns out $T$ is a sphere. This can be seen by induction: The base case is after all of the surgeries. Our manifold is a collection of spheres. The inductive step is a single reverse surgery: by hypothesis will connect a two spheres together, which is always a sphere.
Thus there is a loop $\gamma$ which doesn't bound a disk in $T$ that does bound a disk in $\mathbb{R}^3$. This is the loop OP wanted.
Remark: This is the general proof that no surface $S\subset S^3$ is incompressible 
A: This answer carries through the details of Lee Mosher's comment.
Your goal is to find some loop in the torus that you can do surgery on and get an embedded $S^2$. Equivalently, there is a loop in the torus that bounds an embedded disc in $S^3$ that doesn't interset the torus. (These are equivalent because you can thicken this embedded disc and take its boundary to obtain the $S^1 \times I$ you'll swap out for an $S^0 \times D^2$.). This embedded torus splits the manifold into two sides; call them $M_1$ and $M_2$.
Inspired by the loop theorem, we want to show that there is an embedded loop in $T^2$ such that on one side or the other, it's null-homotopic. Clearly it's null-homotopic in $S^3$. Pick a disc representing that null-homotopy and require that it's transverse to the torus $T^2$ except along its boundary. (You can do this by standard transversality arguments.) Call this map $i$. Pick an innermost circle of $i^{-1}(T^2)$; then because it's innermost, if we restrict $i$ to this circle and the disc it bounds, it now maps only to one of $M_1$ and $M_2$. Suppose for convenience it's $M_1$.
Now that we've got such a loop, we can use the loop theorem to turn the disc it bounds into an embedded one. Unfortunately, I don't really know how to get around using the loop theorem if we're going to do this argument! I could see one trying to prove it for 3-manifolds that live inside $S^3$, but I really don't see how that would help us. But with the loop theorem, we can surger that loop, like you suggest, and invoke Alexander's theorem.
There's also a proof of this that involves basically following along with what Hatcher did for spheres and... doing it for tori. I don't remember the details, but that's how I did this argument when I looked at the book first.
