Having trouble with how to approach this question:
Suppose $G$ is a connected planar graph having girth at least $6.$ Prove that $G$ has at least one vertex with degree at most $2.$
As you initially asked for how to approach this proof:
$(1)$ You can start by assuming $G$ is a connected planar graph having girth at least $6$. Given such a $G$, then, you need to prove $G$ has at least one vertex with degree at most $2$.
$(2)$ You can also prove the statement in its equivalent contrapositive form:
$(3)$ Finally, you can use an indirect proof:
Whatever approach you use in your proof, you'll need to use the definition of a connected planar graph, and one that has girth at least $6$ (where the girth of a graph is the length of its shortest cycle).
You may also need to use Euler's formula as it relates to planar graphs: $$v-e + f = 2,$$ where $v$ is the number of vertices of $G$, $e$ the number of edges of $G$, and $f$ the number of faces. (In planar graphs, the "faces" are the number of regions bounded by edges, including the outer, infinitely large region.)