Basic proof of $d\langle X\cdot M , Y\cdot N\rangle=XY\,d\langle M ,N\rangle$ 
How to proof that: $$\langle X\cdot M , Y\cdot N\rangle_T = \int^T_0 XY d \langle M, N\rangle$$ 
  for $X,Y$ - simple processes and  $M,N \in M^2_c$.

If I need to prove that $$\langle X\cdot M \rangle_T = \int^T_0X^2 d \langle M\rangle$$ then knowing that DM decomposition tells us that:
$X^2 - \langle X\rangle$ is a martingale, I can take conditional expectation and show that: 
$$ \mathbb{E}\left[(X \cdot M)_T^2 - \langle X\cdot M \rangle_T - (X \cdot M)_S^2 + \langle X\cdot M \rangle_S | \mathcal{F}_S\right] = 0$$
What is the way to deal with $\langle X\cdot M , Y\cdot N\rangle_T$?
I suspect that working the same way I will get something similar, but I can not continue:
$$ \mathbb{E}\left[(X \cdot M)_T(Y \cdot N)_T - \langle X\cdot M , Y\cdot N \rangle_T - (X \cdot M)_S(Y \cdot N)_S + \langle X\cdot M, Y\cdot N \rangle_S | \mathcal{F}_S\right] = 0$$
rewriting terms as martingale transforms. Is this a case when we consider simpler case and look at when $X=Y, M=N$?
 A: Hint:


*

*Show (or recall) that the quadratic covariation satisfies $$\langle M,N \rangle_t = \frac{1}{4} \big( \langle M+N \rangle_t - \langle M-N \rangle_t \big) \tag{1}$$ for any two martingales $M,N \in \mathcal{M}_c^2$.

*Use $(1)$ and the fact that $$\langle X \bullet M \rangle_t = \int_0^t X(s)^2 \, d\langle M \rangle_s$$ to show that $$\langle X \bullet M, Y \bullet M \rangle_t = \int_0^t X(s) Y(s) \, d\langle M \rangle_s. \tag{2} $$

*Use $(2)$, $(1)$ and $$\begin{align*} \langle X \bullet M, Y \bullet N \rangle_t &= \frac{1}{4} \bigg( \langle X \bullet (M+N), Y \bullet (M+N) \rangle_t \\ &\qquad -  \langle X \bullet (M-N), Y \bullet (M-N) \rangle_t\bigg) \end{align*}$$ to conclude $$\langle X \bullet M, Y \bullet N \rangle_t = \int_0^t X(s) Y(s) \, d\langle M,N\rangle_s.$$


Alternative approach: Since $X$ and $Y$ are simple processes, the stochastic integrals $X \bullet M$ and $Y \bullet N$ can be calculcated explicitly. Then it is a straight-forward computation to verify that $$(X \bullet M)_t \cdot (Y \bullet N)_t - \int_0^t X(s) Y(s) \, d\langle M,N\rangle_s$$ is a martingale, and this proves the assertion.
