$k$-regular and not complete graph and distance

Question

Let $G$ be a connected, $k$-regular and not complete graph. Then, for any vertex, there exists another vertex with distance two from that.

I am not sure about a rigorous justification for that. I need to get more elaboration about that.

Answer 1

Suppose this is false. Then, take a vertex $A$ where this fails. Then, every vertex is at a distance of $1$ from $A$ as the graph is connected (if there exists any point at greater distance, then there exists a point of distance $2$). Then, there are $k$ other vertices in the graph by construction, which means that we have a $k$ regular graph on $k+1$ vertices, which is a complete graph.

Answer 2

Let $G$ be a connected, $k-$regular graph that is not complete. Suppose for a contradiction that there exists a vertex $v \in V(G)$ such that there are no vertices with distance $2$ to $v$. Notice that if there exists a vertex $u$ with distance more than $2$ to $v$, then we can simply take a vertex from the $uv$ path with distance $2$ to $v$. Then only possibility is every other vertex being distance $1$ to $v$. But then $v$ should be connected to every other vertex. Since $G$ is a regular graph, this means $G$ is complete, which is a contradiction.