Can someone please give a nice proof to the following statement:
Let $G$ be a plane graph and $e$ an edge of $G$. If $e$ lies on a cycle $C \subseteq G$, then $e$ lies on the frontier of exactly two faces of $G$, and these are contained in distinct faces of $C$. If $e$ lies on no cycle, then $e$ lies on the frontier of exactly one face of $G$.
There is a proof in the book of Reinhard Diestel, Graph Theory, but I don't understand it.
I make this up right now, so this is only supposed to be a possible roadmap.
First we note that an edge cannot be incident to more than two faces.
Assume $e$ lies on a cycle. Then the subgraph containing just this cycle splits the plane into two regions (Jordan curve theorem), one bounded one unbounded, and $e$ is incident to both of these. Note that successively reinserting the old edges lying in the bounded region does not decrease the number of faces $e$ is incident to. Indeed, in the process only the bounded region will be subdivided further, which is strictly separated from the unbounded region, so $e$ is incident to the unbounded face and a face inside the bounded region. Now considering the planar graph embedded on a sphere instead of in the plane, we note that the unbounded face can indeed be viewed as a bounded face (by using another projection from sphere to plane). Hence reinserting edges into the unbounded face does not change anything too.
Assume the edge is not contained in a cycle. Then deleting the edge disconnects the graph (further). By using a similar argument to the previous case we can embed all components of the graph in such a way into the plane, such that the endpoints of $e$ lay on the unbounded face. Hence, reinserting $e$ we find that $e$ is incident to the unbounded face only.