Is there any elementary (i.e. without using analytical methods like the theory of Riemann surfaces or more elaborate results from differential geometry) way to show that the universal covering of the compact oriented surface of genus $g>0$ is homeomorphic to $\mathbb R^2$?



Universal covering of compact surfaces
You can build a certain covering space of the surface $S$ rather explicitly as a nested union of closed discs $D\_1 \subset D\_2 \subset D\_3 \subset \cdots$, each contained in the interior of the next, from which it follows that the union is $\mathbb{R}^2$ and, being simply connected, is therefore the universal covering space. To construct these discs, start by representing your surface in the usual fashion as the quotient of a $4g$-gon, with oriented 1-cells assigned labels, and with side pairings respecting orientations and labellings. The disc $D\_1$ is a single copy of the $4g$-gon $Q$ with appropriate side labels, and the map $D\_1 \to S$ is the one given by the quotient map. Then $D\_2$ is constructed by attaching additional copies of $Q$ to the periphery of $D\_2$, exactly as they ought to be: one copy of $Q$ attached to each edge of $D\_1$, and additional copies of $Q$ attached to each vertex so as to fill out $4g$ copies of $Q$ attached around each vertex of $D\_1$. Et cetera.


