This question was answered, and I did this approach. But, I was wondering if a similar approach to the following would work and if not, why:
In our homework, we have show that via Sterographic Projection, that $S^n -\{p\} \cong \mathbb{E}^n$. On this note, I wanted to say that since $[0,1)\times [0,1)$ and $[0,1]\times[0,1)$ are homeomorphic to $D^2$ with an open set on their boundary removed (we have also shown this), that then via Sterographic Projection the two open intervals are homeomorphic, implying the two $S^1$ spheres with an open interval of their boundary removed were homeomorphic, and so the two disks.
Would this approach work? Sorry if it did not totally make sense.
EDIT: I know that a priori, we can't say that because two boundaries are homeomorphic, that then two sets in a topology are homeomorphic. But, I feel like in this case, because the only part where these two shapes that are not homeomorphic is at their boundary should justify the approach I mentioned.
As you say in the edit, this doesn't just work automatically. In fact, it's not even true that any two spaces which are of the form "$D^2$ with an open interval in the boundary removed" are homeomorphic: the open interval you remove might be all of the boundary except one point. You really need to use the fact that the boundary intervals you're removing are not just homeomorphic, but that there is a homeomorphism between them that extends to a homeomorphism $D^2\to D^2$ (and hence also gives a homeomorphism between their complements). Note that it actually suffices to just extend the homeomorphism to a homeomorphism $f:S^1\to S^1$, since you can then extend it to $g:D^2\to D^2$ by defining $g(rv)=rf(v)$ for $r\in [0,1]$, $v\in S^1$.