Wikipedia states that Steinitz's theorem says:
"a given graph $G$ is the graph of a convex three-dimensional polyhedron, if and only if $G$ is planar and $3$-vertex-connected"
So, given a convex polyhedron, there exists a planar graph whose vertices correspond to the vertices of the polyhedron and similar for the edges. However, if one wishes to deduce Euler's formula for polyhedrons from Euler's formula for planar graphs, it is also necessary for the faces of the planar graph (including the unbounded face) to correspond to the faces of the polyhedron.
How can we show that, given a convex polyhedron, a planar graph exists such that, not only the vertices and edges of the graph correspond to those of the polyhedron, but that the faces correspond as well?
I have read in some sources that stereographic projection does the trick, or that a light source can be shined from above the polyhedron, but these arguments are rather intuitive. I would appreciate seeing a proof that is as rigorous as possible or being pointed to one.
Stereographic projection indeed is the trick. Just use the center of projection very Little above one of its faces. Then the whole polyhedron gets projected right into the projection of this very face. This projection then is nothing but the Schlegel diagram of the polyhedron wrt. the chosen face.
Btw. that a polyhedron always provides a graph via its edge skeleton is the easy part. The other way round, i.e. classifying which graphs can be understood in this way as a projection of some polyhedron, that is what the Steinitz theorem is aiming for.
--- rk