The limits $X$ and $Y$ are not unique ... one can show that there always exist independent random variables $U$ and $V$ such that $X \stackrel{d}{=} U$ and $Y \stackrel{d}{=} V$ and $X_n \stackrel{d}{\to} U$, $Y_n \stackrel{d}{\to} V$.
Follows like this: Let $X \sim \mu$, $Y \sim \nu$ for distributions $\mu$, $\nu$. There exists a probability space $(\Omega',\mathcal{A}',\mathbb{P}')$ and independent random variables $U$ and $V$ on $\Omega'$ such that $U \sim \mu$, $V \sim \nu$ (Proof: Heinz Bauer: "Probability Theory" (De Gruyter 1996), Theorem 9.5). Since the convergence in distribution only depends on the distribution of the random variables you still have $X_n \stackrel{d}{\to} U$ and $Y_n \stackrel{d}{\to} V$.
So as far as I can see you don't even need the independence of $X_n$ and $Y_n$ (but you'll need it if you want to proof $(X_n,Y_n) \stackrel{d}{\to} (U,V)$ or $X_n + Y_n \to U+V$).