First I will state the question and then I will show my answer, which I obtained by imposing an additional condition on the processes involved. I would like to get some help on how to solve the original question without this additional condition.
So the question is as follows. Let $M$ and $N$ be continuous local martingales with $M_0 = N_0 = 0$. Suppose that $A$ with $A_0 = 0$ is a continuous adapted finite variation process and that the following holds. $$(M_tN_t - A_t) \quad \text{is a local martingale.}$$ Prove that $A_t = [M,N]_t$ a.s.
My answer makes use of the additional condition that $M$ and $N$ are square-integrable. Then I write $$M_tN_t - A_t = M_tN_t - [M,N]_t + [M,N]_t - A_t$$ The square-integrability of $M$ and $N$ give $MN - [M,N]$ is a local martingale. This implies $[M,N] - A$ is a local martingale. Furthermore, $[M,N]$ is a finite variation process, which combined with $A$ being of finite variation gives $[M,N] - A$ is of finite variation. Since all the processes involved are continuous, so is $[M,N] - A$. For a continuous local martingale $M$ of finite variation, we have $M_t = M_0$. This then implies $$[M,N]_t - A_t = [M,N]_0 - A_0 = 0 \qquad \square$$.
Is it possible to solve this question without the square-integrability condition? I don't want to rule out the possibility that the teacher forgot to mention this condition in the problem statement.
Edit 1: Apparently what I meant with $M$ and $N$ being square-integrable was that they were local $L^2$ martingales. This was my misunderstanding of the $L^2$-ness of local martingales.