if $AB\neq 0$ for any non zero matrix $B$ then $A$ is invertible

Yet another proof. Consider the linear map $L_A : M_n(F)\to M_n(F)$ defined by $$\forall B\in M_n(F)\;:\; L_A(B)=AB\, .$$ The assumption means that $\ker (L_A)=\{ 0\}$, i.e. $L_A$ is 1-1. Since $M_n(F)$ is finite-dimensional, it follows that $L_A$ is invertible. In particular, one can find $B\in M_n(F)$ such that $AB=Id$; so $A$ is invertible.

Remember that $A$ is not invertible iff the equation $Ax=0$ has non-trivial solutions. Can you now construct a non-zero $B$ with $BA=0$?

Suppose $A$ isn't invertible. It follows that $(0,v)$ is an eigenpair of $A$, for some non-null $n\times 1$ vector $v$, that is, $Av=0_{n\times 1}$. Now consider the matrix $B$ whose columns are all equal to $v$.