$\mu_n\rightharpoonup\mu$ and $\nu_n\rightharpoonup\nu\implies\mu_n\otimes\nu_n\rightharpoonup\mu\otimes\nu$. Let $\mu_n,\mu\in P(X)$ and $\nu_n,\nu\in P(Y)$ ($X$ and $Y$ are Polish spaces, or if this doesn't work, maybe $\mathbb{R}^d$). I was trying to reproduce the steps in this answer. Since $\mu_n\rightharpoonup\mu$ and $\nu_n\rightharpoonup\nu$ then $\{\mu_n,\mu\}$ and $\{\nu_n,\nu\}$ are tight sets. Then since $\{\mu_n\otimes\nu_n,\mu\otimes\nu\}$ is also a tight set. Indeed, if we fix $\epsilon>0$ and take compacts $K_{X,\epsilon}, K_{Y,\epsilon}$ with $\mu_n(X\setminus K_{X,\epsilon})<\epsilon, \mu(X\setminus K_{X,\epsilon})<\epsilon, \nu_n(Y\setminus K_{Y,\epsilon})<\epsilon, \nu(Y\setminus K_{Y,\epsilon})<\epsilon$ (which we can do because of tightness of the sets), then $\mu_n\otimes\nu_n(X\times Y\setminus K_{X,\epsilon}\times K_{Y,\epsilon})\le \mu_n\otimes\nu_n(X\setminus K_{X,\epsilon}\times Y\cup X\times Y\setminus K_{Y,\epsilon})\le 2\epsilon$.
Then I would like to show that for every subsequence of $\mu_n\otimes\nu_n$ we have a further subsequence converging weakly to $\mu\otimes\nu$ and I will be done.
Given a subsequence $\mu_{n_k}\otimes\nu_{n_k}$, by tightness, we have a further subsequence that converges to some $\gamma\in P(X\times Y)$, i.e. $\mu_{n_{k_j}}\otimes\nu_{n_{k_j}}\rightharpoonup\gamma$. But I would need some help to show that $\gamma=\mu\otimes\nu$.
edit concerning Olivier Diaz' answer
$\mathcal{U}$ is a $\pi$-system : If, $′×″∈\mathcal{U}$ and $′×″∈\mathcal{U}$, then $′×″\cap ′×″=′\cap ′×″\cap ″$ and $∂(′∩′)⊂∂′∪∂′$ so $(∂(′∩′))≤(∂′)+(∂′)=0$ and similarly $(∂(″∩″))=0$
 A: It is true that if $X$ and $Y$ are separable metric spaces,  $\mu_n$ and $\nu_n$ are sequences Borel probability measures on $X$ and $Y$ respectively, and $\mu_n\Rightarrow\mu$ and $\nu_n\Rightarrow\nu$, then $\mu_n\otimes\nu_n\Rightarrow\mu\otimes\nu$. I don't know if there is an easy way to continue with the approach you outlined, but I doubt it would be easy.
I am aware of a proof of these statement that is based on the Portmanteau theorem.
In Bilinglsley's classic 1968 book on Convergence of Probability Measures, it is  stated in Theorem 3.1, p.p. 20-21
Theorem: If $X$, $Y$ separable metric space, and $P_n$, $P$ are probability measures in $(X\times Y)$ (with the product metric), then as sufficient and necessary condition for $P_n\Rightarrow P$  is that
$$P_n(A'\times A'')\xrightarrow{n\rightarrow\infty}P(A'\times A'')$$
for each $A'$ $\mu$-continuity set, and each $A''$ $\nu$-continuity set, where
$\mu(dx)=P(dx\times Y)$ and $\nu(dy)=P(X\times dy)$ are the corresponding marginals.
In particular, this would hold for product measures $\mu_n\otimes\nu_n$ with $\mu_n\Rightarrow\mu$ and $\nu_n\Rightarrow\nu$.
If you can't access that reference, let me know and I will try to write the main ideas in my answer.
Kallenberg's book Foundations of Probability, 2nd ed. p. 78 has an even more general result:
Theorem: For any sequences of separable metric spaces $S_1,S_2,\ldots$ and random variables $\xi=(\xi^1,\xi^2,\ldots)$, $\xi_n=(\xi^1_n,\xi^2_n,\ldots)$, $n\in\mathbb{N}$, in $\prod_kS_k$, $\xi_n\Rightarrow\xi$ iff for any functions $f_k\in\mathcal{C}_b(S_k)$
$$\mathbb{E}[f_1(\xi^1_n)\cdot\ldots\cdot f_m(\xi^m_n)]\xrightarrow{n\rightarrow\infty}\mathbb[f_1(\xi^1)\cdot\ldots\cdot f_m(\xi^m)]$$
The proof of this theorem is also based on the Portmanteau theorem.

The general ideal is

*

*Notice that $\partial(A'\times A'')\subset\Big((\partial A')\times Y)\Big)\cup \Big(X\times\partial(A'')\Big)$


*Necessity is clear from the Portmanteau. The tricky part is sufficiency. The class $\mathcal{U}$ of sets $A'\times A''$ for which $A'$ and $A''$ are $\mu$-continuity and $\nu$-continuity sets respectively is a multiplicative class: $\partial (B\cap B')\subset\partial(B) \cup \partial(B')$.


*Each set $A'\times A''\in\mathcal{U}$ satisfies $P_n(A'\times A'')\xrightarrow{n\rightarrow\infty}P(A'\times A'')$ by assumption.


*Each open set is countable union of sets in $\mathcal{U}$.


*$\mathcal{U}$ is a determining class. You can approximate any open set by a finite union collection of sets in $\mathcal{U}$ from which you can then show that $\liminf_nP_n(U)\geq P(U)$.


*Portmanteau.

Edit:
Sketch of  proof: (1), (2) clear; (3) is by assumption (we are going to check is a sufficient condition. (4) follows by separability. Every open set is countable union of open boxes whose sides are balls in $X$ and $Y$. The radius can be choses so the they are continuity sets for the marginals (for a fixed $x_0\in X$, there can only be a countable sequence of radius $r$ for which $\mu(\partial B(x_0;r))=P(\partial B(x_0;r)\times Y)>0$.  Similar for a fixed $y_0\in Y$ and the marginal $\nu$.
(5) Suppose $U$ is open and $U=\bigcup_m B_{x_m}\times B_{y_m}$ where $B_{x_n}$ and $B_{y_n}$ are open  continuity balls for $\mu$ and $\nu$ respectively. For $\varepsilon>0$ there is $N$ large enough so that $\mu\big(U\setminus\bigcup^N_{m=1}B_{x_m}\times B_{y_m}\big)<\varepsilon$.
Let $B_m:=B_{x_m}\times B_{y_m}$. Then
$$\begin{align}
P_n\Big(\bigcup^N_{m=1}B_m\Big)&=\sum^N_{m=1}P_n(B_m)-\sum_{1\leq i<j\leq N}P_n(B_i\cap B_j)+\ldots +(-1)^NP_n(B_1\cap\ldots\cap B_N)\\
&\stackrel{n}{\Longrightarrow}\sum^N_{m=1}P(B_m)-\sum_{1\leq i<j\leq N}P(B_i\cap B_j)+\ldots + (-1)^NP(B_1\cap\ldots\cap B_N)\\
&=P\Big(\bigcup^N_{m=1}B_m\Big)>P(U)-\varepsilon
\end{align}$$
Hence $\liminf_nP_n(U)\geq P(U)$. Conclusion follows by Prtmanteau's theorem, i.e. $P_n\stackrel{n}{\Longrightarrow}P$.
