general-topology
Let $X=\{((b+a\cos(\theta))\sin(\phi),(b+a\cos(\theta))\cos(\phi),a\sin(\theta))\in{}\mathbb{R}^3|\theta,\phi\in{}[0,2\pi)\}$ with $0<a<b$.
I am trying to show $X\simeq {}S^1\times{}S^1$.
I understand that if we consider the unit square, with the bottom and top sides identified with identified endpoints, same with the left and right sides, we can embed this in $\mathbb{R}^3$ to get the torus. But can I use this with $X$ to show what I am after?
Modulo showing $\tilde{f}$ is open and the minor corrections given by David in the comments, your proof looks good to me. You could prove that $\tilde{f}$ is open, but in this case there's actually a trick that gives you that $f^{-1}$ is continuous for free without having to think about $\tilde{f}$. Namely, if you can show that $\mathbb{R}^2/\mathbb{Z}^2$ is compact, then $f$ is a continuous bijection from a compact space to a Hausdorff space, so it is automatically a homeomorphism. And to show that $\mathbb{R}^2/\mathbb{Z}^2$ is compact, you can just note that it is the continuous image of the compact space $[0,2\pi]^2$ under the continuous map $\pi$.
From your other question it seems that you've figured out how to show that the quotient is a Mobius strip. For sake of completeness, here's the argument in pictorial form (I apologize for the crude drawings.):
Regarding the last step, maybe this (even more poorly-drawn) picture would help:
Best Answer
Let $R = [0,2\pi] \times [0,2\pi]$. The map $p : [0,2\pi] \to S^1, p(t) = e^{it}$, is a quotient map which identifies the two points $0, 2\pi \in [0,2\pi]$ to a single point in $S^1$.
The map $q = p \times p : R \to T = S^1 \times S^1$ is a quotient map. It identifies $(\theta,0)$ with $(\theta,2\pi)$ and $(0,\phi) $ with $(2\pi,\phi)$ as desribed in your question.
Define $$f : R \to \mathbb R^3, f(\theta,\phi) = ((b+a\cos \theta)\sin \phi,(b+a\cos \theta)\cos \phi,a\sin \theta) .$$ $f(\theta,\phi) = f(\theta',\phi')$ means that
$(b+a\cos \theta)\sin \phi = (b+a\cos \theta')\sin \phi'$
$(b+a\cos \theta)\cos \phi = (b+a\cos \theta')\cos \phi'$
$a\sin \theta = a\sin \theta'$, i.e. $\sin \theta = \sin \theta'$.
Squaring 1. and 2. and adding yields
Since $0 < a < b$, both $b+a\cos \theta, b+a\cos \theta'$ are positive and we conclude
Inserting 5. in 1. and 2. gives us
$\sin \phi = \sin \phi'$
$\cos \phi = \cos \phi'$
We have $\theta, \theta' \in [0,2\pi]$, hence 3. and 5. show that either $\theta = \theta'$ or that one of $\theta, \theta'$ is $0$ and the other is $2\pi$. Similarly, 6. and 7. show that $\phi = \phi'$ or that one of $\phi, \phi'$ is $0$ and the other is $2\pi$.
This shows that $f(\theta,\phi) = f(\theta',\phi')$ if and only if $q(\theta,\phi) = q(\theta',\phi')$. By the universal property of the quotient $f$ induces a unique injective map $f' : T \to \mathbb R^3$ such that $f' q = f$. Since $T$ is compact and $f(R)$ is Hausdorff, we see that $f' : T \to f(R)$ is a homeomorphism. But clearly $$f(R) = f([0,2\pi) \times [0,2\pi)). $$