Let $n$ denote the size of the square matrix $A$. By Theorem NMRRI the hypothesis that $A$ is singular implies that $B$ is not the identity matrix $I_n$. If $B$ has $n$ pivot columns, then it would have to be $I_n$, so $B$ must have fewer than $n$ pivot columns. But the number of nonzero rows in $B$ ($r$) is equal to the number of pivot columns as well. So the $n$ rows of $B$ have fewer than $n$ nonzero rows, and $B$ must contain at least one zero row. By Definition RREF, this row must be at the bottom of $B$.
A proof can also be formulated by first forming the contrapositive of the statement (Proof Technique CP) and proving this statement.