$A$ is an invertible matrix and $B$ is a non-invertible matrix. Can $AB$ be invertible? I have the following idea:
Sup. $AB$ is invertible, then:
$B=IB=(A^{-1}A)B=A^-1(AB)$, then apply inverse both sides:
$B^{-1}=(AB)^{-1}(A^{-1})^{-1}=(AB)^{-1}A$, but $B$ is non-invertible (hip). This leads to a contradiction, as we supposed $AB$ is invertible.
Therefore $AB$ is non-invertible.
I'm not sure if the step where I apply "inverse both sides" is right. Otherwise I'm not sure how to prove this.
Note 1: I CAN'T use $(AB)^{-1}=B^{-1}A^{-1}$ since the hypothesis for that theorem is $A, B$ invertible matrices and this is not the case.
Note 2: I CAN'T use determinants yet.
Your argument is correct, though depending on the level, you may want to explain why $$B^{-1}=\big(A^{-1}(AB)\big)^{-1}=(AB)^{-1}(A^{-1})^{-1}.$$ Another approach would be to suppose that $AB$ is invertible with inverse $C$.Then $$(CA)B=C(AB)=I,$$ so $CA$ is the inverse of $B$, a contradiction.
Alternatively, you could note that if $B$ is non-invertible, then there exists some nonzero vector $x$ such that $Bx=0$. Then also $ABx=0$, and so $AB$ is non-invertible.