= Solution
Put $D=d_2(\mu,\nu)$ and $z_t(x,y)=(1-t)x+ty$. The <marginal distributions> of the optimal <transport plan> $\Pi$ give $\mu_0=\mu$ and $\mu_1=\nu$. Also
$$
\int|z_t|^2\,d\Pi\leq(1-t)\int|x|^2\,d\mu+t\int|y|^2\,d\nu<\infty,
$$
so $\mu_t$ has a finite second <absolute moment>. For $0\leq s<t\leq1$, the <pushforward measure> $(z_s,z_t)_\#\Pi$ is a <coupling> of $\mu_s,\mu_t$. Therefore
$$
d_2(\mu_s,\mu_t)^2\leq\int|z_t-z_s|^2\,d\Pi=(t-s)^2D^2.
$$
Apply the <triangle inequality> for the <second Wasserstein distance> along $\mu_0,\mu_s,\mu_t,\mu_1$:
$$
D\leq d_2(\mu_0,\mu_s)+d_2(\mu_s,\mu_t)+d_2(\mu_t,\mu_1)\leq sD+(t-s)D+(1-t)D=D.
$$
All inequalities are equalities, including when $D=0$. Thus this <displacement interpolation> is a constant speed <metric geodesic>:
$$
\boxed{d_2(\mu_s,\mu_t)=|t-s|d_2(\mu,\nu).}
$$
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