Here we have 2 cases
Case 1: If (u,v) is an edge in the graph then
$d(r,v)=d(r,u)+1$ and hence $d(r,u) \lt d(r,v)$
Case 2: if (u,v) is not an edge, and u and v are children of same parent say
Consider the above tree structure as some part of a given graph where r is my source vertex from which I started my BFS algorithm.
Now, minimum cost to reach X would be $d(r,x)$
Minimum cost to reach U would be $d(r,u)=d(r,x)+1=d(r,v)$ and this is equal to minimum cost to reach v from r via x.
So here we have $d(r,u)=d(r,v)$
Combining cases 1 and 2 we get
$d(r,u) \leq d(r,v)$
Option (C)