Why is $S^1 \times S^1$ a Torus?

I think you're thinking of it in the right way, but you're also trying to have your cake and eat it here. If you want a completely rigorous argument, you can't get around having a rigorous description of the 3d shape that you're trying to prove homeomorphic to $S^1\times S^1$ in the first place. But that means, in particular, that you need a well-defined way of speaking on particular points on the geometrical torus, and how would you do that if not by coordinates?

In fact, one doesn't need that much inspiration to find a concrete coordinate-based homeomorphism. Let's define $S^1$ as the set $\{(x,y)\mid x^2+y^2=1\}\subset \mathbb R^2$ with the subspace topology. Then, $$h((x,y),(z,w)) = z(0,0,1)+(w+2)(x,y,0)$$ is your desired homeomorphism, into the set $$\{(x,y,z)\mid z^2 + \Bigl(2-\sqrt{x^2+y^2}\Bigr)^2 = 1\}\subset \mathbb R^3$$ which one ought to recognize as our geometrical torus.

To describe this as "not very enlightening" would be fair if it was about purely formal algebraic manipulation or black-box recipes for computation -- but in fact it has a clear geometric significance. One might say, instead:

For each $(x,y)$ we draw the $(z,w)$ circle on a vertical plane through the origin and oriented such that it passes through $(x,y)$, with the center of the $(z,w)$ circle being offset by $2$ units in the direction of $(x,y)$.

However, such a verbal description risks getting complex and hard to grasp, and it is all too easy for the writer to introduce ambiguities where it can be understood only if the reader already understands which construction is meant. The coordinate formulas, on the other hand, are unambiguous, and with a bit of practice one can see the geometric construction directly in them.

Well, one way you might find fruitful is to consider $S^1$ as $\mathbb{R}/\mathbb{Z}$. Then if we identify the "square with sides identified" model with $\mathbb{R}^2/\mathbb{Z}^2$, we want to show that $\mathbb{R}/\mathbb{Z}×\mathbb{R}/\mathbb{Z}$ is homeomorphic to $\mathbb{R}^2/\mathbb{Z}^2$, which I find is much more natural since we don't have to treat boundaries as "special cases" when we define the homeomorphism.

Cheers,

Rofler

If you like to think about the torus as the rectangle $[a,b]\times [c,d]$ with the opposite sides identified, then think what happens when you identify the opposite sides.

When you identify $[a,b]\times \{ c \}$ with $[a,b] \times \{ d \}$, every "vertical" segment $\{ x \} \times [c,d]$ becomes a circle $\{ x \} \times S^1$.

Thus, you get the cylinder $[a,b] \times S^1$.

Now, you identify the ends $a \times S^1$ with $b \times S^1$, every "horisontal" segment $[a,b] \times y$ becomes a circle, $S^1 \times y$, where $y$ is a point on $S^1$. Thus you get $S^1 \times S^1$.

P.S. This can also be seen easely by a cutting argument. If you cut $S^1$ at a point you get an interval. So cut $S^1 \times S^1$ twice, first at $\{a \} \times S^1$ to get $[a,b] \times S^1$ and cut it again. Now glue it together backwards and you get your definition of the torus.