Given $X, Y$ independent Brownian motions, I'd like to show that $XY$ is a martingale. This seems to be a fairly easy result, but I can't work it out, nor find anywhere that gives a proper proof.
I can see that it could be done using Ito's formula applied to $f(x,y) = xy$ but I'd like to show it directly from the martingale definition
You didn't specified any filtration... I guess you talk about $\mathcal F_t=\sigma (X_t,Y_t)$.
Hint
Let $s<t$. You have that $$\mathbb E[X_tY_t\mid \mathcal F_s]=\mathbb E[(X_t-X_s)(Y_t-Y_s)\mid \mathcal F_s]+\mathbb E[(Y_t-Y_s)X_s\mid \mathcal F_s]+\mathbb E[(X_t-X_s)Y_s\mid \mathcal F_s]+\mathbb E[X_sY_s\mid \mathcal F_s].$$
Now, using $X_s$ and $Y_s$ are $\mathcal F_s$ measurable, $X_t-X_s$ and $Y_t-Y_s$ are independents and independents of $\mathcal F_s$, the claim follow.