By definition one has $$ W_{\infty}(\mu,\nu)=\inf\limits_{\pi} \|f_d\|_{L^\infty(\pi)}, $$ where the infimum is taken along admissible plans $\pi\in\Pi(\mu,\nu)\subset \mathcal P(X\times x)$ 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 $x,y\in X$ (we work as usual in a Polish space $(X,d)$).
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, $$ d_H(\mathrm{spt}(\mu),\mathrm{spt}(\nu)) \leq d(x_n,y_n) $$ and taking $n\to\infty$ finally gives the result.