We consider $2$ martingales $(X_r)_{r \in \mathbb{R}_+}$ and $(Y_r)_{r \in \mathbb{R}_+}$ (with respect to their canonical filtrations) such that the $2$ processes $(X_r)_{r}$ and $(Y_r)_r$ are independent.
Prove that $(X_rY_r,\sigma(X_u,Y_u,0 \leq u \leq r))$ is a martingale.
Letting $\mathcal{H}_r:=\sigma(X_u,Y_u,0 \leq u \leq r),$ clearly $X_rY_r$ is integrable (by independence) and $\mathcal{H}_r$-measurable.
Considering $u \leq r,$ we need to verify that $E[X_rY_r|\mathcal{H}_u]=X_uY_u$ a.s.
Any ideas how to do it ? Is it true that $\mathcal{H}_u=\sigma(\sigma(X_v,0 \leq v \leq u)\cup\sigma(Y_v, 0 \leq v \leq u))$ so that we use a $\pi$-system argument (by proving that $E[X_rY_r1_{K_1 \cap K_2}]=E[X_uY_u1_{K_1 \cap K_2}]$ for $K_1 \in \sigma(X_v,0 \leq v\leq u),K_2 \in \sigma(Y_v, 0 \leq v \leq u)$)?
Your idea is correct.
Of course. Indeed, $\sigma(X_v,0 \leq v \leq u)\subset\sigma(X_v,Y_v,0 \leq v \leq u)$, and $\sigma(Y_v,0 \leq v \leq u)\subset\sigma(X_v,Y_v,0 \leq v \leq u)$, so $\sigma(X_v,0 \leq v \leq u)\cup\sigma(Y_v, 0 \leq v \leq u)\subset\sigma(X_v,Y_v,0 \leq v \leq u)$, and since $\sigma(X_v,Y_v,0 \leq v \leq u)$ is a $\sigma$-algebra, $\sigma(\sigma(X_v,0 \leq v \leq u)\cup\sigma(Y_v, 0 \leq v \leq u))\subset\sigma(X_v,Y_v,0 \leq v \leq u)$.
Conversely, any set of the form $\{X_v\in A\}$ or $\{Y_v\in B\}$ (for $A$ and $B$ measurable and $v\le u$) belongs obviously to $\sigma(X_v,0 \leq v \leq u)\cup\sigma(Y_v, 0 \leq v \leq u)$ and therefore to $\sigma(\sigma(X_v,0 \leq v \leq u)\cup\sigma(Y_v, 0 \leq v \leq u))$. So $\sigma(\sigma(X_v,0 \leq v \leq u)\cup\sigma(Y_v, 0 \leq v \leq u))$ contains the $\sigma$-algebra generated by those sets, that is $\sigma(X_v,Y_v,0 \leq v \leq u)\subset\sigma(\sigma(X_v,0 \leq v \leq u)\cup\sigma(Y_v, 0 \leq v \leq u))$.
Yes. You can consider the $\pi$-system composed of the sets of the form $K_1\cap K_2$ for $K_1\in\sigma(X_v,0 \leq v\leq u)$ and $K_2\in\sigma(Y_v,0 \leq v\leq u)$.