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Let G be a graph with n vertices and if every vertex has a degree of at least (n−1) / 2 then G is connected.

Proof : Take any two vertex A and B ( suppose n = 3 ) , if A and B is not adjacent then at least (n-1) edges can join them to the remaining vertices. Because both A and B have a degree at least (n-1)/2  .

As total number of vertex is n , if we leave A and B there now (n-2) other vertices exists ..The Pigeonhole principle implies that one of those (n-2) vertices must be adjacent to A and B . Now , every pair of vertex have common neighbor so the graph must be connected. 

Example:

n=3,  so A have degree 1 , B have also degree 1 , if A and B is not adjacent ( means not have a direct edge between them ) then at least 3-1=2 edges can join A and B together.

A     B

 \     /

    C

Diagram 

 

So here in this question,  n = 8  and D = (n-1)/2

so , D =  ⌈  7/2 ⌉   = ⌈ 3.5 ⌉  = 4 

Hence , the minimum value of D must be 4

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considering counter example . Take example of two disjoint K4,K4 and then K4 union K4 (still disconnected) and then +1 to degree.
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