Let $X_n \rightarrow_d X$ and $Y_n \rightarrow_d Y$ where $X$ and $Y$ are i.i.d standard exponential random variables. However, I do not have that for any $n$, $X_n$ and $Y_n$ are independent.
Can I still have the result that for any continuous sets $A$ and $B$:
$$lim_{n \rightarrow \infty} P(X_n \in A \cap Y_n \in B) = P(X \in A)P(Y \in B)? $$
Which conditions on the dependence of $X_n$ and $Y_n$ do I need to claim this?
The comment of @whuber below the OP is crucial, and it brings into the surface a confusion that established notation can create to less experienced users like me.
When we write $X_n \to_d X$ we only mean $F_n(x) \to F(x)$ (these being distribution functions). Namely we only mean that the leading term of the sequence of $X_n$'s, viewed on their own, acquires at the limit a distribution that is the same as the marginal distribution of some $X$ random variable. Same comments go for $Y_n$ and $Y$. But then, the information "$X$ and $Y$ are independent" does not play a role. We only used the symbols $X$ and $Y$ to invoke their marginal distributions, and not their possible statistical association. The fact that $X$ and $Y$ are independent has no bearing on what happens to the limits of $X_n$ and $Y_n$, because the limiting rv's for $X_n$ and $Y_n$ are not necessarily $X$ and $Y$. They may be the random variables $X_n \to_p W_x$ and $Y_n \to_p Z_y$, which may have the marginal distributions of $X$ and $Y$, but they are not $X$ or $Y$. They just have the same marginal distributions. But then, $X$ and $Y$ may be independent while $W_x$ and $Z_y$ are not.
Only if we can claim, making additional assumptions (see https://stats.stackexchange.com/a/379971/28746), that convergence in distribution does imply convergence in probability, then the statistical relation of $X$ and $Y$ becomes relevant.
In such a more narrow case, if we have that $X_n \to_p X$ and $Y_n \to_p Y$, we can say the following:
Assume that $\{X_n\}$ and $\{Y_n\}$ are comprised of continuous random variables. Then their joint distribution function $H_n(x,y)$ has a unique Copula representation
$$H_n(x,y) = C_n\big[F_n(x), G_n(y)\big].$$
Evidently, the Copula is a proper distribution function itself w.r.t $F_n(x), G_n(y)$.
Moreover the limit $C$ of $C_n$ itself will depend on the assumed dependence relation between the limiting random variables. These are assumed independent, so it follows that $C_n$ converges to the Independence Copula $\Pi = F(x)G(y)$. Now, the limiting distribution functions $F(x)$ and $G(y)$ are continuous, so the convergence to them of $F_n(x)$ and $G_n(x)$ is uniform, and distribution functions are bounded. Also the product $F(x)G(y)$ is continuous so the Copula converges uniformly. So we can look at $C_n\big[F_n(x), G_n(y)\big]$ and see a composition of functions that each converges uniformly to its respective limit. It follows that $$\lim C_n\big[F_n(x), G_n(y)\big] = C\big[F(x), G(y)\big] = \Pi[F(x), G(y)] = F(x)G(y).$$
In other words, under convergence in probability, some "regularity/smooth behavior" assumptions are sufficient, together with limiting independence, to obtain
$$H_n(x,y) \to F(x)G(y).$$
