A 4-Regular graph with 7 vertices is non planar A Graph with 7 vertices each having degree 4 cannot be planar. Any hints on the proof?

Show that the graph must contain a $K_{3,3}$ configuration.
Pick any pair of non-adjacent vertices, $v_1$ and $v_2$. Since $v_1$ and $v_2$ each have degree $4$ and there are only $5$ other vertices, they must have at least $3$ common neighbors.
Then, try to find a third vertex $v_3$ adjacent to the same common neighbors, thus constructing $K_{3,3}$.
Edit: Take $v_1$ and $v_2$ as described above. They must have at least $3$ common neighbors (and at most $4$). If they have $4$ common neighbors, then the remaining vertex shares the same $4$ neighbors as $v_1$ and $v_2$, so this forms a $K_{3,3}$ configuration.
So, say $v_1$ and $v_2$ share $v_3,v_4,v_5$ as common neighbors, with $v_1$ adjacent to $v_6$ and $v_2$ adjacent to $v_7$.
If $v_6$ and $v_7$ are not adjacent, then they each share $v_3,v_4,v_5$ as common neighbors with $v_1$ and $v_2$, giving a $K_{3,3}$ configuration.
If $v_6$ and $v_7$ are adjacent, then they are each adjacent to exactly two of $v_3,v_4,v_5$, and furthermore, they cannot be adjacent to the same pair. So say $v_6$ is adjacent to $v_3,v_4$ and $v_7$ is adjacent to $v_4,v_5$. Then, we have a $K_{3,3}$ configuration made of $v_1,v_2,v_6$ and $v_3,v_4,v_5$, where the 'edge' connecting $v_6$ to $v_5$ goes through $v_7$.
