Show that $(X_{n},Y) \to^{\mathcal{D}} (X,Y)$ AND if $X=h(Y)$ where $h$ is a Borel function that $X_{n}\to^{P} X$ 
Let $X_{n}$, $X$, and $Y$ be real-valued r.v.'s all defined on the same space $(\Omega, \mathcal{A},\mathbb P)$. Assume that $\lim_{n \to \infty}\mathbb E\{f(X_{n})g(Y)\}=\mathbb E\{f(X)g(Y)\}$ whenever $f$ and $g$ are bounded, and $f$ is continuous, and $g$ is Borel. Show that the sequence $(X_{n}, Y)$ converges in law to $(X,Y)$. If furthermore $X=h(Y)$ for some Borel function $h$, show that $X_{n}\to^{\mathbb P} X$.

I have made several attempts at proving the first part, and have failed (see comments below). The second part I am not sure how to prove either.  
 A: The proof of the convergence in distribution of $(X_n,Y)$ to $(X,Y)$ is done here (the use of characteristic functions could make it shorter).
Assume that $X=h(Y)$ for some Borel function $h$. Using the assumption with $g:=f\circ h$, a bounded Borel-measurable function, we obtain that for each continuous and bounded function $f$, 
$$\lim_{n\to \infty}\mathbb E[f(X_n)f(X)]=\mathbb E[f(X)^2].$$
Noticing that $f^2$ is a continuous bounded function, we obtain that 
$$\tag{*}\lim_{n\to \infty}\mathbb E[(f(X_n)-f(X))^2]=0.$$
Let $\varepsilon\gt 0$. For a fixed $R$, we have 
$$\mathbb P\{|X_n-X|\gt\varepsilon\}\leqslant \mu\{|X_n|\gt R\}+\mu\{|X|\gt R\}+\mathbb P\{|X_n-X|\gt\varepsilon, |X_n|\leqslant R,|X|\leqslant R\}.$$
Consider $f$ a continuous bounded function such that $f(x)=x$ on $[-R,R]$. Then 
$$\mathbb P\{|X_n-X|\gt\varepsilon, |X_n|\leqslant R,|X|\leqslant R\}\leqslant \varepsilon^{-2}\mathbb E[(f(X_n)-f(X))^2\chi(|X_n|\leqslant R,|X|\leqslant R\})]\leqslant \varepsilon^{-2}\mathbb E[(f(X_n)-f(X))^2],$$
hence for each $R$
$$\limsup_{n\to +\infty}\mathbb P\{|X_n-X|\gt\varepsilon\}\leqslant \sup_n\mu\{|X_n|\gt R\}+\mu\{|X|\gt R\}.$$
Using the fact that $X_n\to X$ in distribution, we obtain that the RHS goes to $0$ as $R$ goes to infinity. 
