$A$ is a $n\times n$ matrix, whose $n$ row vectors are all different. then, we can get rid of one column of $A$(there exist a column, we abandon this column ), such that the new $n\times (n-1)$ matrix $B$ also has $n$ different row vectors.
With Mathematical Induction, we can prove the statement. But, Is any other methods?
[Editor's note: Below is my interpretation of what the question is meant to be (based on the exchange of comments below), JL]
Given an $n\times n$ matrix $A$ with distinct rows show that we can select one of its columns in such a way that removing the said column yields an $n\times (n-1)$ matrix $B$ that still has distinct rows.
Apparently the OP knows of a way of proving this by induction, and is looking for a different approach.
What you are saying is not true. For example, take the matrix $$A=\begin{bmatrix}1 & 0\\ 1 & 1\end{bmatrix}$$ which has $2$ row vectors which are different. Now, get rid of the second column of $A$. Then the matrix $$B=\begin{bmatrix}1\\1\end{bmatrix}$$ which you get does not have $2$ different row vectors.