Method 1:
If all vertices have different degrees, then the degree sequence will be {1,2,3,....n-1}, it will not have ‘n’( A simple graph will not have edge to itself, so it can have edges with all other (n-1) vertices). Degree sequence has only (n-1) numbers, but we have ‘n’ vertices. So, by Pigeonhole principle there are two vertices which has same degree.
Method 2:
A) consider a triangle, all vertices has same degree, so it is false
C) consider a square with one diagonal, there are less than three vertices with same degree, so it is false
D) consider a square with one diagonal, vertices have different degrees. So, it is false.
We can conclude that option B is correct.