Prove that $AB=BA=0$ for two idempotent matrices.

Suppose that $A, B$ are idempotent matrices ($A^2=A$), such that $A + B$ is idempotent, prove that $AB = BA = 0$

This is true in every ring where you can divide by $2$. So it is true in particular for matrices.

We have $(A+B)^2=A^2+AB+BA+B^2=A+AB+BA+B$. Since $A+B$ is idempotent, it follows that $$AB+BA=0.$$ We call this equation (E).

Right-multiply (E) by $B$. We find $AB+BAB=0$, hence $AB=-BAB$.

Left-multiply (E) by $B$. This yields $BAB+BA=0$, hence $BA=-BAB$.

Equating the last two equations, we find $AB=BA$.

Now $AB+BA=0$ clearly yields $2AB=2BA=0$. Dividing by $2$, we get $$ AB=BA=0. $$

This is incorrect.

For example take $F=\mathbb{F}_{2}$ the field with two elements and $A=B=I$ over $F$.

$A,B$ are clearly idempotent and $$A+B=I+I=2I=0$$ hence $$(A+B)^{2}=0^{2}=0=A+B$$ is also idempotent.

But $$AB=BA=I^{2}=I\neq0$$

From your assumption you can only get $AB+BA=0$.