We define the Torus by $T^2 = S^1\times S^1$. Another way to look at it is to pick the square $[0,1]\times [0,1]$ and identify opposite sides.
Indeed, if we identify the points $(0,y)$ with $(1,y)$ we get a cylinder, while if we identify the points $(x,0)$ and $(x,1)$ also we get the torus intuitively.
Now, I want to show that with this equivalence relation $[0,1]\times [0,1]$ is actually homemoprhic to $T^2=S^1\times S^1$.
My guess is as follows: we can define $f : [0,1]\times [0,1]\to T^2$ by
$$f(x,y)=(e^{2\pi ix},e^{2\pi i y}).$$
Now, $f$ is surjective and is continuous by properties of the product topology because each entry is continuous. Other than that $f$ is a class function. Indeed we can check that:
If $(x,y)\in (0,1)\times (0,1)$ then $[(x,y)] = \{(x,y)\}$ and there's nothing to be shown.
For each $y\in [0,1]$ we have $[(0,y)]=[(1,y)]=\{(0,y), (1,y)\}$. Here we can actually show explicitly that $f(0,y)=f(1,y)$ because $\exp(2\pi i x)$ is $2\pi$-periodic.
For each $x\in [0,1]$ we have $[(x,0)]=[(x,1)]=\{(x,0),(x,1)\}$. The proof that $f(x,0)=f(x,1)$ is analogous to the preceding case.
Hence $f$ descends to a map $\varphi : ([0,1]\times [0,1]/\sim) \to T^2$ which is continuous and injective. Since $f$ was also surjective, $\varphi$ also is, and hence $\varphi$ is a continuous bijection.
On the other hand, $[0,1]\times [0,1]$ is compact and so $[0,1]\times[0,1]/\sim$ is compact on the quotient topology.
Also, since $S^1\times S^1 \subset \mathbb{R}^4$ and $\mathbb{R}^4$ is Hausdorff, it follows $T^2$ is Hausdorff.
Since $\varphi$ is a continuous bijection from a compact space to a Hausdorff space, it follows $\varphi$ is a homeomorphism.
My question here is: is this proof correct? Or is there something wrong with it? Is there some way I could improve this proof and make it better?