Proof that the optimal transport cost is lower semicontinuous In Chapter 6 of Villani's Optimal Transport: Old and New, it is stated that

... the Wasserstein distance is lower semicontinuous on $P(\mathcal{X})$ (just like the optimal transport cost $C$, for any lower semicontinuous cost function $c$; recall the proof of Theorem 4.1).

I am trying to verify this claim. The best idea I have is as follows. Suppose we have measures $\mu_k \to \mu$ and $\nu_k \to \nu$ weakly. Let $\pi^\ast \in \Pi(\mu,\nu)$ and $\pi^\ast_k \in \Pi(\mu_k, \nu_k)$ be optimal couplings for $c$, which exist by Theorem 4.1 (assuming appropriate bounding conditions, e.g. nonnegativity). If we have $\pi^\ast_k \to \pi^\ast$ weakly, then we can use the lower semicontinuity of the cost functional (see Lemma 4.3 in Villani) to conclude
$$C(\mu, \nu) = \int c \, d\pi^\ast \leq \liminf_k \, \int c \, d\pi^\ast_k = \liminf_k \, C(\mu_k, \nu_k).$$
Only problem is, I'm not sure whether $\pi^\ast_k \to \pi^\ast$ weakly -- I can't prove it. Is this the case?
One thing that makes me question this approach is that I'm not sure how it follows the proof of Theorem 4.1. Can anyone see this?

Addition: I found a proof of this claim in Lemma 5.2 of this paper, but I do not understand it. I reproduce it here in my notation, and with some additional exposition.
Suppose we have measures $\mu_k \to \mu_0$ and $\nu_k \to \nu_0$ weakly on a Polish space $\mathcal{X}$ as $1 \leq k \to \infty$. Then the sets $M := \{\mu_k\}_{k \geq 0}$ and $N := \{\nu_k\}_{k \geq 0}$, being convergent sequences containing their limit points, are both compact in the weak topology, and are hence tight by Prokhorov's theorem. As such, the set $\Pi(M, N)$ of all couplings whose marginals lie in $M$ and $N$ respectively is tight, by Lemma 4.4 of Villani.
Let $\pi^\ast_k$ be an optimal coupling of $\mu_k$ and $\nu_k$. Since each $\pi^\ast_k \in \Pi(\mu_k, \nu_k) \subseteq \Pi(M, N)$, which is tight, we have that $\pi^\ast_{k'} \to \pi$ for some subsequence $\pi^\ast_{k'}$, and for some $\pi \in P(\mathcal{X} \times \mathcal{X})$ (i.e. the set of probability measures on the product space). 
We will show that $\pi \in \Pi(\mu_0, \nu_0)$. For any bounded continuous $f : \mathcal{X} \to \mathbb{R}$, we see that
$$\int_{\mathcal{X} \times \mathcal{X}} f(x) \, d\pi^\ast_{k'}(x, y)
\to \int_{\mathcal{X} \times \mathcal{X}} f(x) \, d\pi(x,y)
= \int_{\mathcal{X}} f \, d\pi(\cdot \times \mathcal{X}).
$$
But likewise
$$\int_{\mathcal{X} \times \mathcal{X}} f(x) \, d\pi^\ast_{k'}(x, y)
= \int_\mathcal{X} f \, d\mu_{k'}
\to \int_{\mathcal{X}} f \, d\mu_0.
$$
Consequently, $\pi(\cdot \times \mathcal{X}) = \mu$, by the answer to this question, since $f$ was arbitrary bounded continuous. Similarly we can show $\pi(\mathcal{X} \times \cdot) = \nu$.
We can then proceed as follows:
$$
C(\mu_0, \nu_0)
\leq \int c \, d\pi
\leq \liminf_{k' \to \infty} \int c \, d\pi^\ast_{k'}
= \liminf_{k' \to \infty} C(\mu_{k'}, \nu_{k'}),
$$
where the first inequality holds because $C(\mu_0, \nu_0)$ is defined as an infimum over all couplings in $\Pi(\mu_0, \nu_0)$. The proof I am reproducing essentially stops at this point.
This all seems promising, but what we really want is
$$
C(\mu_0, \nu_0)
\leq \liminf_{k \to \infty} C(\mu_{k}, \nu_{k}),
$$
and it isn't clear to me that we can infer this from the fact that it holds for our convergent subsequence. In general, for a sequence $x_k$ with subsequence $x_{k'}$, I believe we have
$$
\liminf_k x_k \leq \liminf_{k'} x_{k'}.
$$
What is going on here?
 A: I only refer to the statement of your problem (I did not read the "addition" part).
The sequence of optimal $\pi_k$ is contained in the $w^*$-compact set $\Pi({\mu},{\nu})$ (Villani proves its compactness in the proof of Theorem 4.1 using Prokhorov's theorem and Lemma 4.4) therefore has a converging subsequence. Now just take this subsequence instead of $\pi_k$ and proceed as you mentioned.
This is exactly what Villani does in the proof of Theorem 4.1, hence the reference.
A: Consider a sequence $(\mu_k,\nu_k)\rightharpoonup(\mu,\nu)$. We have to show
\begin{equation}
\liminf_{k\to\infty} C(\mu_k,\nu_k)\ge C(\mu,\nu)
\end{equation}
Consider a subsequence $(\mu_{k_j},\nu_{k_j})$ that converges to the inferior limit, i.e.
\begin{equation}
C(\mu_{k_j},\nu_{k_j})\to \liminf_{k\to\infty} C(\mu_k,\nu_k)\text{ when }j\to \infty 
\end{equation}
Now consider $\pi_k$ to be optimal plans between $\mu_k$ and $\nu_k$
Note that since by definition $\mu_{k_j}\rightharpoonup\mu$ and $\nu_{k_j}\rightharpoonup\nu$ we have that that $P=\{\mu_{k_j},\mu\}$ and $Q=\{\nu_{k_j},\nu\}$ are tight, so by lemma 4.4 of Villani we get that $\Pi(P,Q)$ is tight. however $(\pi_{k_j})_{j\in\mathbb{N}}\subset \Pi(P,Q)$ so by Prokhorovs theorem, we get a further converging subsequence and a plan $\pi\in\mathcal{P}(\mathbb{R}^d\times\mathbb{R}^d)$ such that
\begin{equation}
\pi_{{k_j}_{\ell}}\rightharpoonup\pi\text{ when }\ell\to\infty
\end{equation}
Now by the stability of optimal transport (proposition 5.20 from Villani) we know that $\pi$ is actually an optimal plan between $\mu$ and $\nu$ and by lower semi-continuity of cost functionals (lemma 4.3 from Villani) we get
\begin{equation}
\liminf_{\ell\to\infty}C(\mu_{{k_j}_{\ell}},\nu_{{k_j}_{\ell}})=\liminf_{\ell\to\infty}\int c(x,y)d\pi_{{k_j}_{\ell}}(x,y)\ge\int c(x,y) d\pi(x,y)=C(\mu,\nu)
\end{equation}
However, by definition
\begin{equation}
\liminf_{\ell\to\infty}C(\mu_{{k_j}_{\ell}},\nu_{{k_j}_l})=\lim\limits_{j\to\infty} C(\mu_{k_j},\nu_{k_j})=\liminf_{k\to\infty} C(\mu_k,\nu_k)
\end{equation}
which concludes the proof.
