I "proved" that if $A_{m\times n},B_{n\times m}$ are matrices such that $\text{rank}(A)=\text{rank}(B)=m$ then $ABA\neq 0$. The problem is that the statement isn't true because $\begin{pmatrix}0 & 1 \end{pmatrix}\begin{pmatrix}1 \\ 0\end{pmatrix}\begin{pmatrix}0 & 1 \end{pmatrix}=\begin{pmatrix}0 & 0 \end{pmatrix}$. Here is the "proof":
Since $A$ is of order $m\times n$ then there exists a linear transformation $T$ from $F^n$ to $F^m$ such that $A$ is the transformation matrix of $T$. For the same reason there exists a linear transformation $S$ from $F^m$ to $F^n$ such that $B$ is the transformation matrix of $S$. Observe that $m=\text{rank}(B)=\text{dim}(\text{Im}S)$. We know that $\text{dim}(\text{Ker}(S))+\text{dim}(\text{Im}(S))=\text{dim}(F^m)=m$ and we conclude that $\text{dim}(\text{Ker}(S))=0$ which means that $\text{Ker}(S)=\{0\}$.
Consider the linear Transformation $R$ from $F^n$ to $F^n$ such that for every $v\in F^n$ we define $R(v)=S(T(v))$ or in other words $R=ST$. Because $\text{Ker}(S)=\{0\}$ then in particular $\text{Ker}(R)=\{0\}$ and therefore $R$ is an isomorphism. Since $R$ is an isomorphism we know that the transformation matrix of $R$ is invertible, but that matrix is $BA$. Now if we assume that $ABA=0$ then we get $A=0$ which contradicts $\text{rank}(A)=m>0$.
Where did I go wrong?
This is false.
$\ker S=\{0\}$ doesn't lead to $\ker R=\{0\}$. Because
$$Rv=0\Leftrightarrow STv=0\Leftrightarrow Tv=0\color{red}\nRightarrow v=0$$
$v=0$ if and only if $\ker T=\{0\}$, but there is no way to derive this conclusion. For example, you give the counter-example, $T=\begin{pmatrix}0 & 1 \end{pmatrix},~ S=\begin{pmatrix}1 \\ 0\end{pmatrix}$, with this choice, $$\ker T=\text{span}\{(1,0)^T\}\neq\{0\}$$