Every planar graph has a vertex of degree at most 5.

Prove it by contradiction. Suppose that there exists a planar graph with all vertices having degree at least $6$. Then by the handshake lemma, $E = 3V$. If $G$ is planar, then $E \leq 3V - 6$, a contradiction.

Assume that every vertex of a planar graph $G$ has degree $6$ or more. Thus by The First Theorem of Graph Theory we have $2m\geq 6n$ where $n$ and $m$ denote the order and size of $G$. Since $G$ is planar we know that $m\leq3n-6$ and we have $3n\leq m$. Thus $3n\leq 3n-6$ which implies that $0\leq -6$ which is a contradiction. Thus every planar graph has a vertex of degree at most $5$.