TITLE: Degeneracy of outerplanar graphs
QUESTION [9 upvotes]: Does anyone know an elegant proof to the fact that every outerplanar graph has a vertex of degree at most 2 (and hence is 2-degenerate, since every subgraph is also outerplanar). 
I have a proof by induction (on the number of vertices) in mind, but it is long and somewhat cumbersome (it splits into a few cases). Can anyone point me to a more elegant proof? 
Maybe one can do it using the dual graph - if the dual graph is circle-free then we are done, but I couldn't find an easy argument for that either.
Thanks in advance for any help.

REPLY [5 votes]: Observe that every outerplanar graph can be made into a maximal outerplanar graph of the same order. The regions in the interior of a maximal outer planar graph form a tree since if there was a cycle, that would surround a vertex, contradicting outerplanarity. Trees have at least two leaves. Any region corresponding to a leaf will have a vertex of degree 2. This vertex must have had degree less than or equal to 2 in the original graph.