[proofplan]
The hypotheses give exactly the narrow convergence and moment convergence needed to upgrade each sequence to convergence in the Wasserstein metric. Applying the characterization of $W_p$-convergence, we obtain $W_p(\mu_k,\mu)\to 0$ and $W_p(\nu_k,\nu)\to 0$. The desired convergence then follows from the triangle inequality for the metric $W_p$, applied twice to compare $W_p(\mu_k,\nu_k)$ with $W_p(\mu,\nu)$.
[/proofplan]
[step:Upgrade narrow convergence and moment convergence to Wasserstein convergence]
Because $\mu_k \to \mu$ narrowly and
\begin{align*}
\int_X d(x,x_0)^p\,d\mu_k(x) \to \int_X d(x,x_0)^p\,d\mu(x),
\end{align*}
the characterization of Wasserstein convergence by narrow convergence and convergence of $p$-moments applies to the sequence $(\mu_k)_{k \in \mathbb N}$ and gives
\begin{align*}
W_p(\mu_k,\mu) \to 0.
\end{align*}
Likewise, because $\nu_k \to \nu$ narrowly and
\begin{align*}
\int_X d(x,x_0)^p\,d\nu_k(x) \to \int_X d(x,x_0)^p\,d\nu(x),
\end{align*}
the same characterization applies to the sequence $(\nu_k)_{k \in \mathbb N}$ and gives
\begin{align*}
W_p(\nu_k,\nu) \to 0.
\end{align*}
Here we are using the characterization theorem for $W_p$-convergence on $\mathcal P_p(X)$: narrow convergence together with convergence of the $p$-moment about one, equivalently every, base point is equivalent to convergence in $W_p$ (citing a result not yet in the wiki: Characterization of Wasserstein convergence by narrow convergence and convergence of p-moments).
[guided]
We first convert the stated hypotheses into the form that interacts directly with the Wasserstein metric. The relevant result is the characterization of $W_p$-convergence on $\mathcal P_p(X)$: for probability measures in $\mathcal P_p(X)$ on a Polish [metric space](/page/Metric%20Space), convergence in $W_p$ is equivalent to narrow convergence plus convergence of the $p$-moments about one, equivalently every, base point.
For the sequence $(\mu_k)_{k \in \mathbb N}$, the theorem's hypotheses give both required inputs. First, $\mu_k \to \mu$ narrowly. Second, for the chosen point $x_0 \in X$,
\begin{align*}
\int_X d(x,x_0)^p\,d\mu_k(x) \to \int_X d(x,x_0)^p\,d\mu(x).
\end{align*}
Since every $\mu_k$ and $\mu$ lies in $\mathcal P_p(X)$, the measures have finite $p$-moment, so the characterization theorem applies. Therefore
\begin{align*}
W_p(\mu_k,\mu) \to 0.
\end{align*}
The same verification applies to the sequence $(\nu_k)_{k \in \mathbb N}$. The hypotheses give $\nu_k \to \nu$ narrowly and
\begin{align*}
\int_X d(x,x_0)^p\,d\nu_k(x) \to \int_X d(x,x_0)^p\,d\nu(x).
\end{align*}
Since every $\nu_k$ and $\nu$ lies in $\mathcal P_p(X)$, the characterization theorem again applies, and we obtain
\begin{align*}
W_p(\nu_k,\nu) \to 0.
\end{align*}
This is the only place where the moment convergence assumption is used. Narrow convergence alone would not control how mass may escape to infinity, while convergence of the $p$-moments supplies exactly the missing tightness at order $p$.
[/guided]
[/step]
[step:Compare the two Wasserstein distances by the triangle inequality]
Since $W_p$ is a metric on $\mathcal P_p(X)$, its triangle inequality gives
\begin{align*}
W_p(\mu_k,\nu_k) \le W_p(\mu_k,\mu) + W_p(\mu,\nu) + W_p(\nu,\nu_k).
\end{align*}
Using the symmetry of $W_p$, this becomes
\begin{align*}
W_p(\mu_k,\nu_k) - W_p(\mu,\nu) \le W_p(\mu_k,\mu) + W_p(\nu_k,\nu).
\end{align*}
Applying the same triangle inequality with the roles of $(\mu_k,\nu_k)$ and $(\mu,\nu)$ interchanged gives
\begin{align*}
W_p(\mu,\nu) \le W_p(\mu,\mu_k) + W_p(\mu_k,\nu_k) + W_p(\nu_k,\nu).
\end{align*}
Again using symmetry, we obtain
\begin{align*}
W_p(\mu,\nu) - W_p(\mu_k,\nu_k) \le W_p(\mu_k,\mu) + W_p(\nu_k,\nu).
\end{align*}
Combining the two one-sided estimates yields
\begin{align*}
\left|W_p(\mu_k,\nu_k)-W_p(\mu,\nu)\right| \le W_p(\mu_k,\mu) + W_p(\nu_k,\nu).
\end{align*}
[/step]
[step:Pass to the limit in the comparison estimate]
From the first step,
\begin{align*}
W_p(\mu_k,\mu) \to 0
\end{align*}
and
\begin{align*}
W_p(\nu_k,\nu) \to 0.
\end{align*}
Therefore the right-hand side of
\begin{align*}
\left|W_p(\mu_k,\nu_k)-W_p(\mu,\nu)\right| \le W_p(\mu_k,\mu) + W_p(\nu_k,\nu)
\end{align*}
converges to $0$. Since the left-hand side is non-negative for every $k \in \mathbb N$, the squeeze principle gives
\begin{align*}
\left|W_p(\mu_k,\nu_k)-W_p(\mu,\nu)\right| \to 0.
\end{align*}
Equivalently,
\begin{align*}
W_p(\mu_k,\nu_k) \to W_p(\mu,\nu).
\end{align*}
This is the desired stability statement.
[/step]
