Note that this is equivalent to $$m-r \geq s-q, $$ the difference between rows and rank of $A$ is greater than in $B$. Now, suppose $A: k^n\to k^m$ where $k$ is your favourite field. Let $U\subseteq k^m$ be the subspace given by the first $s$ coordinates. Then $im(B)=U\cap im(A)$ under the natural embedding.
Let $V=im(A)+U$, then from the second isomorphism theorem we have $$V/im(A) \cong U/im(B) $$ and upon taking dimensions we get $$dim(V) -r = s-q. $$ The inequality follows.
Edit: Here is a simpler solution,
rearrange to get $$r-q\leq m-s. $$ So the difference in rank is less than the difference in rows. There are $r-q$ rows that add the the rank of $A$ but not to the rank of $B$. These rows cannot be in the $s$ selected rows since then this would increase the rank of $B$. Therefore, they are one of the $m-s$ rows not in $B$. The inequality follows.