**EDIT**: answer 2 below is completely false, as pointed out by the OP. However this is such a typical example of wishful thinking that I believe it is worth leaving for the posterity. (I'll record it below as False Answer 2.) The right line of reasoning goes the other way around (take quasi-optimizers in the Hausdorff distance instead of taking a quasi-optimal plan), and actually provides finer information.

**Preliminaries**:
I'll consider as usual the case of a Polish space $(X,d)$) with cost function $c(x,y)=d(x,y)$, and I'll also write $Y=X$ for notational convenience.
By definition one has
$$
W_{\infty}(\mu,\nu)=\inf\limits_{\pi} \|f_d\|_{L^\infty(\pi)},
$$
where the infimum is taken along admissible plans $\pi\in\Pi(\mu,\nu)\subset \mathcal P(X\times Y)$ with first and second marginals $\mu,\nu$, respectively, and $f_d$ is just a short notation for $f_d(x,y)=d(x,y)$, the distance function as a function of two variables $x,y\in X$.


----------

**Correct answer 1**




We only consider the case when $D= d_H(\mathrm{spt}\mu,\mathrm{spt}\nu)>0$ (otherwise the statement is trivial).
In words, the proof goes as follows: by definition there are points $x\in\mathrm{spt}\mu$ or $y\in\mathrm{spt}\nu$ which must be at distance at least $D$ from the other support. Hence any admissible transportation plan must eventually ship some (positive amount of) mass from $\mathrm{spt}\mu$ to $\mathrm{spt}\nu$ over a distance $D$. This immediately gives the result, since the infinity-Wasserstein distance just corresponds to taking the best such plan.

More precisely, recall that by definition the Hausdorff distance is
$$
d_H(\mathrm{spt}\mu,\mathrm{spt}\nu)=\max\Bigg\{
\sup\limits_{x\in \mathrm{spt}\mu} d(x,\mathrm{spt}\nu) , \sup\limits_{y\in \mathrm{spt}\nu} d(y,\mathrm{spt}\mu)
\Bigg\}.
$$
Let us assume that the maximum is
$$
D=d_H(\mathrm{spt}\mu,\mathrm{spt}\nu)=\sup\limits_{x\in \mathrm{spt}\mu} d(x,\mathrm{spt}\nu)
$$
(the other case is completely symmetric). Picking a maximizing sequence $x_n\in \mathrm{spt}\mu$, we have
$$
D_n:=d(x_n,\mathrm{spt}\nu)\to D.
$$
Since $x_n\in \mathrm{spt}\mu$, by definition of the support of a measure, we have $\mu(B_\epsilon(x))>0$ for all $\epsilon >0$, as small as desired.
Taking in particular $\epsilon_n =\frac 1n D_n\to 0$ we see by triangular inequality that any point $x\in B_\epsilon(x_n)$ is at distance at least $(1-\frac 1n)D_n$,
\begin{equation}
d(x,y)\geq (1-\frac 1n)D_n,\qquad
\forall x\in B_{\epsilon_n}(x_n)\text{ and }\forall\,y\in \mathrm{spt}\nu.
\tag{1}
\end{equation}
This strongly suggests that the $L^\infty$ norm of $f_d=d(\cdot,\cdot)$ should be at least $(1-\frac 1n)D_n\approx D$, but some extra care is needed to check that the set $B_{\epsilon_n}(x_n)\times\mathrm{spt}\nu $ is not negligible.
To this end, take any admissible plan $\pi\in \Pi(\mu,\nu)$, and recall that any such plan actually satisfies $\mathrm{spt}\pi\subset \mathrm{spt}\mu\times \mathrm{spt}\nu$. By definition of the marginals we have
$$
\pi(B_{\epsilon_n}(x_n)\times \mathrm{spt}\nu)
=
\pi(B_{\epsilon_n}(x_n)\times Y)
=
\mu(B_{\epsilon_n}(x_n))>0.
$$
As a consequence, and from (1),
$$
\|f_d\|_{L^\infty(\pi)}\geq (1-\frac 1n)D_n.
$$
Taking $n\to\infty$ we see that
$$
\|f_d\|_{L^\infty(\pi)}\geq D=d_H(\mathrm{spt}\mu,\mathrm{spt}\nu)
\qquad 
\forall\,\pi\in\Pi(\mu,\nu)
$$
and the conclusion follows.
Note that we actually proved a stronger result, namely that the Hausdorff distance controls the infinity-tansportation for any admissible plan.

----------
**False answer 2**


Let $\pi_n$ be minimizing sequence. By definition of the essential supremum, for any such fixed $n$ there exists $(x_n,y_n)\in \mathrm{spt}(\pi_n)$ such that
$$
d(x_n,y_n)=f_d(x_n,y_n)\geq \|f_d\|_{L^\infty(\pi_n)}-\frac 1n.
$$
Since also $d(x_n,y_n)\leq \|f_d\|_{L^\infty(\pi_n)}$, and because $\pi_n$ is a minimizing sequence, we conclude that
$$
d(x_n,y_n)\rightarrow W_\infty(\mu,\nu)
$$
Recall now that, for any admissible plan $\pi\in \Pi(\mu,\nu)$, one has $\mathrm{spt}(\pi)\subset\mathrm{spt}(\mu)\times \mathrm{spt}(\nu)$.
The previous inequality immediately gives, by definition of the Hausdorff distance,
$$
d_H(\mathrm{spt}(\mu),\mathrm{spt}(\nu))
\leq d(x_n,y_n)
$$
**this is blatantly false!!!**
and taking $n\to\infty$ finally gives the result.