Question on the proof of a property of the rank of a matrix The task is that I have to prove this statement:

Given $(m+1)\times(n+1)$ matrix $A$ like this:
  $$
A=\left[\begin{array}{c |cc}
  1 & \begin{array}{ccc}0 & \cdots & 0\end{array} \\
\hline
  \begin{array}{c}0\\ \vdots\\0\end{array} & {\Large B} \\
\end{array}\right]
$$
where $B$ is a $m\times n$ sub-matrix of $A$.
Show: $\operatorname{rank}(A) = r$  implies $\operatorname{rank}(B) = r - 1$.

Here are the steps that I constructed:
(1) First I already proved this claim: if $m\times n$ matrix $S$ has rank $r$, then $r\le m$ and $r\le n$. And I let $\operatorname{rank}(B) = k$.
(2) Use (1), I say the following 2 statements:


*

*for matrix $A$, $r\le m+1$ and $r\le n+1$.  Thus $r-1\le m$ and $r-1 \le n$

*for matrix $B$, $k\le m$ and $k\le n$.


(3) Now, this is the part that I feel shaky about.  I plan to say that using (2), I have $r-1\le k$ and $k\le r - 1$.  Thus $r - 1 = k$.  But I'm not certain whether this argument is valid.  After noticing that $r - 1\le m$ and $k \le m$ at a same time, I come up with the above relation between $k$ and $r - 1$.
Would someone please help me check if there is anything wrong or missing in this proof?
Thank you.

In analogy, you're arguing: "$5 \leq 7, 4 \leq 7 \Rightarrow 5 \leq 4$", which, when put this way, is clearly not true. 
The way to prove this exercise depends on how your class introduced these concepts. Absent that knowledge, I would argue as follows:  
When reduced to row-echelon form, you can read off the rank of a matrix by the number of non-zero rows. So reduce your matrix $B$ to $1 \leq k \leq \min(n, m)$ non-zero rows (which equals the rank of $B$), and note that the elementary matrices multiplied to the left and right of $A$, to achieve this, will not change column 1 (this is quite obvious as all but the first element of column $1$ and row $1$ are $0$). So you still have the $1$ in position $a_{11}$ (and $0$ in the other entries of the first column), and the entire matrix has $k+1$ non-zero rows in row-echelon form which equals the rank of $A$. 
So $\operatorname{rank}(A) = \operatorname{rank}(B)+1,$ which was to be shown.
