Call $A,B$ the two supports, and $h$ the Hausdorff distance. Given a transport plan $\gamma$ we need to infer $h(A,B)\leq\sup_\gamma d(x,y)$, that is, for any $x\in A$, that $\inf_{y\in B} d(x,y)\leq m$ for all $m\in\mathbb R$ such that $\gamma(\{(x,y):d(x,y)>m\})=0$. This is equivalent to $\exists y_n\in B : \limsup d(x,y_n)\leq m$. Assume it is false. By contradiction, $\forall y_n\in B$ we get $\limsup d(x,y_n)> m$ which implies $d(x,y)>m$ for all $y\in B$. So $0=\gamma(\{x\}\times B)=\nu(B)$, contradiction.

Call $A,B$ the two supports, and $h$ the Hausdorff distance. Given a transport plan $\gamma$ we need to infer $h(A,B)\leq\sup^\gamma d(x,y)$, that is, for any $x\in A$, that $\inf_{y\in B} d(x,y)\leq m$ for all $m\in\mathbb R$ such that $\gamma(\{(x,y):d(x,y)>m\})=0$. This is equivalent to $\exists y_n\in B : \limsup d(x,y_n)\leq m$. Assume it is false. By contradiction, $\forall y_n\in B$ we get $\limsup d(x,y_n)> m$ which implies $d(x,y)>m$ for all $y\in B$. So $0=\gamma(\{x\}\times B)=\nu(B)$, contradiction.