Showing AB=0 does not imply either A,B=0, but that singular

Your proof is good. A product $AB$ can be the zero matrix with $A$ being invertible (or non-singular): just take $B=0$.

Your assignment is to prove that from $AB=0$ it follows that one among $A$ and $B$ is singular.

Now, if $A$ is invertible, then $AB=0$ implies $B=A^{-1}AB=A^{-1}0=0$, so $B$ is certainly singular. QED

Determinants are surely not needed for this.

You can prove more: if $AB=0$ and both $A$ and $B$ are non zero, then both are singular. Indeed, take a nonzero column of $B$, call it $v$; then $Av=0$ and so $A$ is singular. Then apply the same to $B^TA^T$, showing that $B^T$ is singular.

There's nothing wrong with your proof. It can be argued to be even "more basic" than the determinant one.

No need for a contradiction. Matrix $C$ is singular iff $Cx=0$ for some nonzero $x$. $AB=0$ implies that for any $x\neq 0$, $ABx=0$. If $Bx=0$, you are done and $B$ is singular. If $Bx=y\neq 0$, then $Ay=ABx=0$ and $A$ is singular.

With determinants: $AB=0$ implies $\det(AB)=\det(A)\det(B)=0$, so either $\det(A)=0$ or $\det(B)=0$ or both.