Is the unit circle homeomorphic to the unit square? Let $S_1:=\{(x,y\in\mathbb{R}^2:x^2+y^2=1\}$ and $S_2:=\{x,y\in\mathbb{R}^2:\max\{|x|,|y|\}=1\}$ be endowed with subspace topologies induced by the usual topology on $\mathbb{R}^2$.
Then open sets in $S_1$ and $S_2$ are unions of open intervals in $S_1$ and $S_2$ respectively.
Partition $S_1$ into four arbitrary non-empty disjoint open sets $A,B,C,D$ and four distinct points, $a,b,c,d$. 
Similarly, partition $S_2$ into four disjoint open sets $A',B',C',D'$ and four points $(1,1), (1,-1), (-1,-1), (-1,1)$, where $A'$ is the right vertical side of the square, $B'$ is the bottom side of the square, etc.
Let $f:S_1\rightarrow S_2$ be a function such that $f(A):=A',f(B):=B',f(C):=C',f(D):=D'$, and $f(a):=(1,1),f(b):=(1,1),f(c):=(1,1),f(d):=(1,1)$. 
It seems that $f$ is bijective and continuous, and $f^{-1}$ is also continuous, but I can't give a rigorous proof.
 A: If you want to do it rigorously, you will have to define $f$ rigorously. To do that, I suggest you you construct a function that uses the fact that every line starting at $(0,0)$ and going into one direction has exactly one intersection with $S_1$ and exactly one intersection with $S_2$. So, when you have a point $(x,y)$ on $S_2$, you look at the line from $(0,0)$ to that point, and look at where the line intersects $S_1$. You then map $(x,y)$ to that intersection.
It should be fairly easy to show that in that case, $(x,y)$ maps to $$\frac{(x,y)}{\sqrt{x^2+y^2}}$$

Alternatively, you can use the fact that
$$f:[0,2\pi]\to S_1\\
f(t)=(\cos t, \sin t)$$
is a surjective continuous function and that $f(0)=f(2\pi)$.
You can now take pieces of this function to draw parts of the circle. For example, if you limit $f$ to $[0,\frac{\pi}{2}]$, the image of $f$ is exactly one quarter of the circle. So, you can use this to map one side of the square to one quarter of the circle, and since there are $4$ sides to a square and $4$ quarters to a circle, that should be enough to make a homeomorphism.
