The co-variation is the unique continuous process of finite variation such that $XY -[X,Y]$ is a martingale. It is therefore enough to verify that $XY$ is indeed a martingale w.r.t. the filtration $\mathscr F(X) \vee \mathscr F(Y)$ generated by $X$ and $Y$. But this follows from properties the conditional expectation: The independence of $\sigma(X_t) \vee \mathscr F(X)_s$ and $\sigma(Y_t) \vee \mathscr F(Y)_s$ yields $$ E(X_tY_t | \mathscr F(X)_s \vee \mathscr F(Y)_s) = E(X_t | \mathscr F(X)_s ) E(Y_t | \mathscr F(Y)_s )=X_s Y_s.$$

The co-variation is the unique continuous process of finite variation such that $XY -[X,Y]$ is a martingale. It is therefore enough to verify that $XY$ is indeed a martingale w.r.t. the filtration $\mathscr F(X) \vee \mathscr F(Y)$ generated by $X$ and $Y$. But this follows from properties the conditional expectation: The independence of $\sigma(X_t) \vee \mathscr F(X)_s$ and $\sigma(Y_t) \vee \mathscr F(Y)_s$ yields $$ E(X_tY_t | \mathscr F(X)_s \vee \mathscr F(Y)_s) = E(X_t | \mathscr F(X)_s ) E(Y_t | \mathscr F(Y)_s )=X_s Y_s.$$

EDIT: This argument works for continuous independent martingales.