How do you prove to any two simplyconnected domains in the plane are homeomorphic without using the Riemann mapping theorem? An elementary proof would be nice.

$\begingroup$ Such a proof would certainly need the Jordan curve theorem (at least for smooth simple plane curves), and you might not call that elementary. $\endgroup$ – John Pardon May 26 '11 at 13:12

$\begingroup$ @JohnPardon why does it necessarily need Jordan theorem? $\endgroup$ – Fedor Petrov Jun 25 '15 at 18:15

$\begingroup$ The Schoenflies theorem isn't particularly hard to prove. There's a nice cutandpaste style argument given in Hatcher's 3manifolds notes. $\endgroup$ – Ryan Budney Jun 25 '15 at 23:34
Here's how one proof goes. I'm omitting some details, at least for now.
If $U$ is a simplyconnected domain in the plane $\mathbb C$, then define the function $r: U \to \mathbb R_+$ to be the maximum radius of a (round) open disk in the standard Euclidean metric whose interior is contained in $U$. If $r(z) = \infty$ for any $z$, then $U$ is the entire plane, so there is nothing to prove. Otherwise, define a new Riemannian metric using the conformal factor $1/r$, that is, $ds' = 1/r\ ds$ where $ds$ is arc length in the standard metric.
Note that if $U$ is the upper halfplane, the metric coincides with the hyperbolic metric. If $U = \mathbb C \setminus \{0\}$, then all complex linear automorphisms of $\mathbb C$ preserve the metric, and the metric is that of a cylinder that is parametrized isometrically by $\log(z) / \left \langle 2 \pi i \right \rangle$. If $U$ is a round disk, then $ds'$ is a smooth negativelycurved metric except at the center of the disk, where there is a nonsmooth point, but no cone angle: if the disk has radius 1, then a circle about the center of radius $\epsilon$ has length $2 \pi \epsilon / (1\epsilon)$
In general, although the metric $ds'$ need not be smooth, it always has nonpositive curvature. Intuitively: the $1/r$ factor means it takes infinite arc length to reach the boundary, and shortest $ds'$ geodesics try to thread their way into any bays and inlets of $U$ keeping far from the shoreline, since the speed limit is drastically reduced near the shore. In particular, there is a unique $ds'$ geodesic between any two points in $U$, and geodesics have the unique continuation property, they are determined by the tangent vector at the beginning and the length.
To parametrize $U$ by $\mathbb R^2$, choose any point $z_0$ in $U$. The tangent space to $U$ at $z_0$ parametrizes $U$, by $V$ goes to the geodesic through $V$ whose length is the length of $V$.
Sorry for leaving off details. Somewhere else on MO I believe I posted an alternate way to do this, using the convex hull of $S^2 \setminus U'$, where $U'$ is the stereographic image of $U$ on a sphere; in the projective hyperbolic metric, this boundary of the convex hull always is isometric to the hyperbolic plane, from which a proof is easy.
Since you ask for an elementary argument, let's assume that the domains have polygonal boundary. Then, the solution to the "carpenter's rule problem" (by ConnellyDemaineRote and Streinu) gives an actual algorithm for transfroming the domains into convex domains, for which any number of arguments work. (the carpenter's rule problem is described here:
Let me try to solve it completely elementary. My aim is to represent simplyconnected 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 $f_k:P_k\rightarrow \{z:z\leq 12^{k}\}$, with $f_k(\partial P_k)=\{z:z=12^{k}\}$, 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 $P_1,\dots,P_{n1}$ are constructed, consider the square $S_n$ and denote $F=P_{n1}\cup S_n$. 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 $P_{n1}\subset {\rm Int}(F)$ and put $P_n:=F$. Our construction garantees that $\cup P_i=\cup S_i=\Omega$ as desired.