Hint In any planar graph with 6 vertices you have
$$e \leq 3v-6=12 .$$
This proves that any graph with at least 13 edges is non-planar.
All you need to do now, is check the graphs with 12 or less edges.
Since $K_5$ has $10$ edges and $K_{3,3}$ has $9$ edges, any non-planar graph has at least 9 edges.
9 edges: Your graph must be $K_{3,3,}$ (Why?)
10 edges: Your graph must be $K_{3,3,}$ with an extra edge or $K_5$ with an isolated vertex. (Why?)
11 edges and 12 edges: If your graph contains $K_{3,3}$, it is easy: it must be $K_{3,3}$ with enough extra edges.
For $K_5$ you need to look at two different situations: your graph contains $K_5$ as a subgraph, or the sixth vertex is the vertex of degree $2$ in a subdivision of $K_5$.
As for 1), it seems you got the right idea but what does "Every two vertices include adjacent ones as well." mean ?
My personal way of saying it would be : if you have two vertices $u, v$ at distance 3 or 5,
then necessarily there's a vertex at distance 2 from $u$ on the shortest $u - v$ path (That's what point 2. says !).
So the only available odd distance is actually 1, which leaves only one possibility.
For 2), you might want to complete the proof by stating why you can't have $d(u, v_i) < i$, nor $d(u, v_i) > i$ for any $1 \leq i < k$. That is, $d(u, v_k) = k$ by definition, but what about the other values of $i$.
Best Answer
Hint: If $d(u_0,v)=n\gt0$ then $u_0$ has a neighbor $u_1$ such that $d(u_1,v)=n-1$.