Let $X = \mathbb{R}$ and suppose that the probability measure $\pi\in P(X^2\times X^2)$ has the marginals $\delta_x\times\delta_y$ and $\delta_x\times\nu$, where $x, y \in X$ and $\nu$ is a probability measure on $X$. Can we conclude that $\pi$ is the product of its marginals, that is, $\pi=\delta_x\times\delta_y\times\delta_x\times\nu$?
Proposition 1 (see e.g. here). If $\pi \in P(Y \times Y)$ has marginals $\delta_x$ and some probability measure $\mu$, then $\pi = \delta_x \times \mu$.
I tried to mimic the proof of proposition 1 to show the question posed above can be answered affirmatively:
Proof. We want to show that for measurable subsets $A, B, C, D \subset X$, we have $$ \pi(A \times B \times C \times D) = \delta_x(A) \nu(B) \delta_x(C) \delta_y(D), $$ which would yield $\pi=\delta_x\times\delta_y\times\delta_x\times\nu$, since rectangles are a $\cap$-stable generator of the $\sigma$-algebra on $X \times X$.
Case 1. Suppose that $x \not\in A$. Then $$ 0 \le \pi(A \times B \times C \times D) \le \pi(A \times B \times X \times X) = \delta_x(A) \nu(B) = 0 $$ and thus $\pi(A \times B \times C \times D) = 0 = \delta_x(A) \nu(B)$.
Case 2 Suppose $x \in A$, then $$ \delta_x(A) \nu(B) \delta_x(C) \delta_y(D) = \nu(B) \delta_x(C) \delta_y(D) = \nu(B) \pi(X \times X \times C \times D) \ge \nu(B) \pi(A \times X \times C \times D). $$ We now want to show $\delta_x(C) \delta_y(D) \le \pi(A \times X \times C \times D)$, but I haven't been able to.
If $x \not\in C$ or $y \not\in D$, then both sides are zero. If $x \in C$ and $y \in D$, then the left side is 1, so we would have to show $\pi(A \times X \times C \times D) \ge 1$, but I am stuck there.
If we could prove that we would have $$ \delta_x(A) \nu(B) \delta_x(C) \delta_y(D) = \nu(B) \delta_x(C) \delta_y(D) \le \pi(A \times X \times C \times D) \le \delta_x(A) \nu(B) \delta_x(C) \delta_y(D), $$ which yields the statement. $\square$
Can this proof be completed or is the statement even false?
Since $\delta_{\{x\}} \times \delta_{\{y\}} = \delta_{\{ (x, y) \}}$ is Dirac measure on $X^2$, we can use proposition 1 with $Y = X^2$ to conclude that $\pi$ is indeed the product of its marginals.