Let $\mathcal{P}_{X,Y}$ be a collection of joint probability measure of $(X,Y) \in \mathbb{R}^2$. Suppose that $\mathcal{P}_{X,Y}$ is sequentially compact in the topology of weak convergence.
My question: Given a marginal distribution $P_X$, is the following set sequentially compact?
\begin{align} \mathcal{D}_{X,Y,P_X}=\{ P_{XY} \in \mathcal{P}_{X,Y}: P_X \text{ is a marginal of } P_{XY}\}. \end{align}
In other, we fixing a marginal distribution.
My argument: If $\mathcal{P}_{X,Y}$ is sequentially compact than any sequence $ \{P_{X_n,Y_n}\}_{n=1}^\infty$ has a subsequence that converges in $\mathcal{P}_{X,Y}$. This also implies to the sequence $ \{P_{X,Y_n}\}_{n=1}^\infty$ where the marginal is fixed. That is a sequnce $\{P_{X,Y_n}\}_{n=1}^\infty$ has a convergent subsquence $\{P_{X,Y_n}\}_{n_k=1}^\infty$ that convergence in $\mathcal{P}_{X,Y}$, but every elemetn of $\{P_{X,Y_n}\}_{n_k=1}^\infty$ is also in $\mathcal{D}_{X,Y,P_X}$. So, this subsequence convergence in $\mathcal{D}_{X,Y,P_X}$. Therefore, $\mathcal{D}_{X,Y,P_X}$ is sequentially compact.
This is a correct argument?
A more direct argument is the following, based on the idea that Nate Eldridge mentions in his comment: A probability measure $P$ has margin $Q$ if, for all bounded continuous $f:\mathbb R\to\mathbb R$, we have $P(f\otimes1) = \iint_{\mathbb R^2} f(x)P(dx,dy) = Q(f) = \int_{\mathbb R} f(x) Q(dx)$, where $f\otimes1:(x,y)\mapsto f(x)$. Thus $$\{P:P\text{ has margin }Q \}=\bigcap_{f\in C_b(\mathbb R)} \{ P:P(f\otimes1) = Q(f)\}.$$ But for each $f$ the map $P\mapsto P(f\otimes1)$ is continuous, so the set of $P$ such that $P(f\otimes1)=Q(f)$ is closed. So the set of $P$ with margin $Q$ is the intersection of closed sets, and hence closed.
Finally, if $K$ is a compact set of probability measures on $\mathbb R^2$, the subset of $K$ of measures with margin $Q$ is a closed subset of $K$, and hence also compact.