The answer is $\boxed{4}$: Construction: Consider the following graph where the blue vertex has two neighbors and the ten green vertices have three neighbors. Notice that the distance in between the blue vertex and the top green vertex on the inner layer is $4$. Now we verify this graph satisfies the condition involving sub-graphs $A$. The case $|A|=2$ is obvious. Now assume $2< |A|\leq 4$. If their exists vertices from $A$ that lie on both pentagons then there will be at least two boundary points belonging to each pentagon hence $A$ has at least $4$ neighbors. The cases were all the vertices of $A$ lie on one pentagon can be easily verified. [asy][asy] draw((0,2)--(2,0)); draw((2,0)--(1,-2)); draw((1,-2)--(-1,-2)); draw((-1,-2)--(-2,0)); draw((-2,0)--(0,2)); draw((0,4)--(4,0)); draw((4,0)--(2,-4)); draw((2,-4)--(-2,-4)); draw((-2,-4)--(-4,0)); draw((-4,0)--(0,4)); draw((0,2)--(0,4)); draw((2,0)--(4,0)); draw((1,-2)--(2,-4)); draw((-1,-2)--(-2,-4)); draw((-2,0)--(-4,0)); filldraw(circle((0, 2), 0.2), green, red); filldraw(circle((2, 0), 0.2), green, red); filldraw(circle((-2, 0), 0.2), green, red); filldraw(circle((1, -2), 0.2), green, red); filldraw(circle((-1, -2), 0.2), green, red); filldraw(circle((0, 4), 0.2), green, red); filldraw(circle((4, 0), 0.2), green, red); filldraw(circle((-4, 0), 0.2), green, red); filldraw(circle((2, -4), 0.2), green, red); filldraw(circle((-2, -4), 0.2), green, red); filldraw(circle((0, -4), 0.2), blue, red); [/asy][/asy] Bound: We will prove that from each vertex of degree $3$ we can reach all other vertices in at most $4$ moves. In at most $1$ move we can reach $4$ vertices. By the conditions of the problem, in at most $2$ moves we can reach more than $8$ vertices. Then it is not hard to show that it will take it most $2$ moves to reach all of the remaining vertices.

redacted,read wrong the statement

Answers above are incorrect , you just proved that for any proper graph $G$ , there exist two vertices with distance $3$ and also the answer is actually $4$. For constructing a proper graph with diameter $4$ , consider a circle with vertices $v_1 , v_2 , v_3 , v_4 , v_5 , v_6$ and another one with vertices $w_1 , w_2 , w_3 , w_4 , w_5$ and also connect pairs $v_1 , w_1$ and $v_2 , w_2$ and $v_3 , w_3$ and $v_5 , w_4$ and $v_6 , w_5$.

wait, do we have to find the maximum or the minimum diameter ? @below: sorry

motannoir wrote: wait, do we have to find the maximum or the minimum diameter ? The problem statement asks for the maximum!

Shayan-TayefehIR wrote: motannoir wrote: wait, do we have to find the maximum or the minimum diameter ? The problem statement asks for the maximum! Understood, when it was posted before it asked for the minimum I will fix my solution.