Prove that there exists a simple path of length $1$ or $2$ from $u$ to $v$

Question

Let $G=(V,E)$ be a simple graph with $|V|=17$ and $\forall v\in V\quad \deg(v)\geq 8$.
Prove that for every $v,u \in V$ there exists a simple path of length $1$ or $2$, which starts with $u$ and ends with $v$.

Progress:
By contradiction, assume there exist $u,v\in V$ for every simple path from $u$ to $v$ (let it be $S$) with $3\leq |S|$ or $0=|S|$. I don't see what I can do here, since $0=|S|$ looks true.

Tried building a simple path with no success as well.

Answer 1

You have to assume the graph is simple. Then $u$ connects to at least $8$ other vertices, leaving at most $8$ that it does not connect to. If $v$ is among the vertices $u$ connects to, we are done. Otherwise $v$ connects to at least $8$ other vertices. There are only seven others that do not connect to $u$, so it must connect to one of the vertices that does connect to $u$.

Answer 2

The closed neighborhoods of $u$ and $v$ both contain $\geq 9$ vertices. Since the number of vertices is $17 < 2 \times 9$, these two closed neighborhoods intersect.

Note: We need to also assume $u \neq v$ for the theorem to be correct.