= Solution
Let $D=\operatorname{diam}(X)<\infty$, and first suppose $1\leq p\leq q<\infty$. On any <transport plan> $\pi$, which has total mass one, the <Holder inequality> gives
$$
\left(\int|x-y|^p\,d\pi\right)^{1/p}
\leq\left(\int|x-y|^q\,d\pi\right)^{1/q}.
$$
Also, $|x-y|\leq D$ gives
$$
\int|x-y|^q\,d\pi\leq D^{q-p}\int|x-y|^p\,d\pi.
$$
Taking infima, or using arbitrarily close competitors if an infimum is not attained, yields
$$
\boxed{W_p(\mu,\nu)\leq W_q(\mu,\nu)
\leq D^{1-p/q}W_p(\mu,\nu)^{p/q}.}
$$
For $p=q$ the two distances coincide, and if $D=0$ there is only one <probability measure>, so the conclusion is immediate. For $D>0$ and $p<q$, the displayed bounds imply both directions of
$$
\boxed{W_p(\mu_n,\mu)\to0\quad\Longleftrightarrow\quad W_q(\mu_n,\mu)\to0.}
$$
For $q<p$, exchange the exponents. Thus all finite-order <p-Wasserstein distances> induce the same convergence on a bounded subset of $\mathbb R^d$. Closedness of $X$ is not required for these estimates.
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