 Solution

Let $D=\mathrm{diam}(X)<\infty$, and first suppose $1\le p\le q<\infty$. On any transport plan $\pi$, which has total mass one, the Holder inequality givesAlso, $|x-y|\le D$ givesTaking infima, or using arbitrarily close competitors if an infimum is not attained, yieldsFor $p=q$ 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 ofFor $q<p$, 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.