Let $X_n\to$ $X$ in distribution, $Y_n\to Y$ in distribution, and assume they are all independent. Does that imply $X_nY_n\to XY$ in distribution?
My idea: From Durrett's book, if $Y=c$, a constant, then the above is true even without independence.
Now since they are independent, my idea is to use characteristic functions. But I have no idea how to work with $$ E(e^{itX_nY_n}). $$
Yes.
You can show $(X_n,Y_n)\Rightarrow(X,Y)$: the $(X_n,Y_n)$ are jointly tight, and any limit law must be (1) a product measure and (2) have the desired margins. Then appeal to the continuous mapping theorem: the function $(x,y)\mapsto xy$ is continuous.
Alternatively, you can use a Skorohod coupling: construct $X_n'$ with the same distribution as $X_n$, and $X'$ with the same distribution as $X$, for which $X_n'\to X'$ almost surely, and similarly for $Y_n'$ etc. Then $X_n'Y_n'\to X'Y'$ almost surely, so the result falls out.
Or, I suppose, a $2\epsilon$ argument based on Fubini's theorem can make your characteristic function argument work: $E\exp(itX_nY_n) = E\phi_{Y_n}(tX_n)$ which is close to $E\phi_Y(tX_n)$ which is close to $E\phi_Y(tX)=E\exp(itXY)$, and so on. I have not thought through all the error bounds here, but clearly the dominated convergence theorem will be very useful.
Or you could note that the finite sums of bounded continuous functions of form $f(x)g(y)$ are dense in the space of all continuous bounded functions of two variables, so $Ef(X_n)g(Y_n)\to Ef(X)g(Y)$ (which follows by independence) for all continuous bounded $f$ and $g$ implies $Eh(X_n,Y_n)\to Eh(X,Y)$ for all continuous bounded $h$.
It is well known that weak convergence implies uniform convergence of characteristic functions on compact sets. We have $Ee^{ityX_n} \to Ee^{iytX}$ uniformly for y in compact sets. Given $\epsilon >0$ let T be such that $P\{|Y_n|>T\}<\epsilon$ for all n. This is possible by tightness. Now consider $\int |Ee^{ityX_n} - Ee^{iytX}| d\mu _n (y)$. Split the integral into the one over $[-T,T]$ and the one over its complement. By uniform convergence on $[-T,T]$ it follows quite easily that $\int |Ee^{ityX_n} - Ee^{iytX}| d\mu _n (y) \to 0$ which says $Ee^{itX_n Y_n}-Ee^{itXY_n} \to 0$. The fact that $Ee^{itX Y_n}-Ee^{itXY} \to 0$ follows from the fact that $Ee^{itX Y_n}-Ee^{itXY} \to 0$ 'boundedly ' and $ \int |Ee^{itX Y_n}-Ee^{itXY}|d\mu (x) \to 0$ by Dominated Convergence Theorem, $\mu$ being the distribution of X. Combining the two limit relations we get $Ee^{itX_n Y_n}-Ee^{itXY} \to 0$.
