TITLE: Riemann mapping theorem for homeomorphisms
QUESTION [24 upvotes]: How do you prove to any two simply-connected domains in the plane are homeomorphic without using the Riemann mapping theorem? An elementary proof would be nice.

REPLY [9 votes]: Let me try to solve it completely elementary. My aim is to represent simply-connected domain $\Omega$ as a union $\cup_{i=1}^{\infty} P_i$, where $P_1,P_2,\dots$ are (closed) polygons and $P_i\subset {\rm Int}(P_{i+1})$ ($\rm Int$ is for interior). If such representation is obtained, then we may construct homeomorphisms $f_k:P_k\rightarrow \{z:|z|\leq 1-2^{-k}\}$, with $f_k(\partial P_k)=\{z:|z|=1-2^{-k}\}$, and $f_{k+1}$ extends $f_k$. For $x\in \Omega$ define $f(x)=f_k(x)$ for all $k$ such that $x\in P_k$, we get a homeomorphism from $\Omega$ to the open unit disc $\{z<1\}$.
How to construct our system of polygons? Enumerate all diadic squares $[a\cdot 2^n,(a+1)\cdot 2^n]\times [b\cdot 2^n,(b+1)\cdot 2^n]$, $a,b,n$ are integer, containing in $\Omega$: $S_1,S_2,\dots$. Define $P_1=S_1$. Next, if polygons $P_1,\dots,P_{n-1}$ are constructed, consider the square $S_n$ and denote $F=P_{n-1}\cup S_n$. If necessary, add to $F$ finite number of dyadic squares so that it becomes connected. It may contain holes, but since $\Omega$ is simply connected everything in holes belongs to $\Omega$, add all this to $F$. Finally, add several small dyadic cells to $F$ so that $P_{n-1}\subset {\rm Int}(F)$ and put $P_n:=F$. Our construction garantees that $\cup P_i=\cup S_i=\Omega$ as desired.