This is a homework problem and my solution. I think I get the main ideas and understand what is going on, but I need to work on my proof writing. I don't get full credit when I think I should get a little better than what I have been getting.
Theorem: For $n \geq 2$, planar graphs with $n$ vertices have at least two vertices whose degree is at most 5.
Proof: Let $G$ be a planar graph. We will do cases, $n=2$ then $n > 2$.
Case 1. For $n=2$, if $G$ is a planar graph with $n$ vertices then either $G$ is the complete graph of order $2$, $K_2$, or $G$ is the null-graph of order $2$, $N_2$. In the former case each vertex has degree 1 and in the latter case each vertex has degree 0 so the condition is satisfied.
Case 2. Suppose $n >2$. We will further assume that $G$ is connected. Then since $G$ is a planar graph, we have $$e \leq 3n -6$$ where $e$ is the number of edges of $G$.
Assume for contradiction that at most one vertex has degree smaller than or equal to 5.
Suppose there is only one such vertex, $v_k$ where $\mathrm{deg}(v_k)\leq 5$. Then by the handshaking lemma we have $$2e = \sum_{i=1}^n \mathrm{deg}(x_i) \geq 6(n-1) + \mathrm{deg}(v_k)$$ $$\Rightarrow \mathrm{deg}(v_k) \leq 2e-6n+6 \leq 2(3n-6)-6n+6=-6$$ a contradiction.
Suppose there are no such vertices, then all vertices has degree larger than or equal to 6.
Then,
$$2e = \sum_{i=1}^n \mathrm{deg}(x_i) \geq 6n \Rightarrow e \geq 3n$$ $$\Rightarrow 3n \leq e \leq 3n -6 $$ $$\Rightarrow 0 \leq -6$$ a contradiction.
Suppose that $G$ is disconnected and has connected components $G_1,G_2,...G_m$. If any of the connected components have $n \geq 2$ vertices, the condition is satisfied by above. The other case is they all only have $1$ vertices in which case $G$ is just a null graph and the condition is satisfied.
This is really more of a draft but it has all my ideas.
Thank you very much if you took the time to read this.