Prove that every outer planar graph is $3$-choosable
Here is what I think the proof look like
I will prove this by induction on the order of $G$
Base: $n=3$ then $\Delta(G) \leq 2$, thus $\chi_l(G) \leq 1+ \Delta(G) \leq 3$, hence $G$ is $3$-choosable
Inductive step: Assume that this is true for every outer planar graph of order $n-1$. Since $G$ is outer planar, $G$ contain $2$ vertices of degree $2$ or less, so let $v$ be one of these vertice, $G-v$ is $3$-choosable by the hypothesis of inductive step. Since $deg(v) \leq 2$, $v$ is adjacent to at most $2$ vertices, so for every collection of color list of size $3$, we can always color $v$ properly.
I keep getting comment that I didn't define all variable that I used in my proof. So I wonder if anyone would please check if my proof is correct, or there is anything that I should define but I didn't.
Best Answer
Your proof is fine. Note the family is "outerplanar" graphs not "outer planar" graphs.
And my wife (Joan Hutchinson) tells me that a stronger results applies here:
if two vertices have 2-lists and all others 3-lists, then the graph is list-colorable as shown in J.P. Hutchinson, On list-coloring extendable outerplanar graphs, Ars Math. Contemp 5 (2012) 171-184, available at amc.imfm.si
But the 2-lists cannot be reduced even to a 1-list and a 2-list or three vertices with 2-lists.