Let $(X,d_X)$ and $(Y,d_Y)$ be compact metric spaces ($X,Y$ with at least two points each to avoid triviality) and $f:X\times Y\to\mathbb{R}$ a Lipschitz function, that is, there is $C>0$ such that for all $x,x'\in X$, $y,y'\in Y$ $$|f(x,y)-f(x',y')|\leq C(d_X(x,x')+d_Y(y,y'))\ .$$ Define for each $x\in X$ the zero marginal superlevel set of $f$ (for the lack of a better name...) $$F(x)\doteq f(x,\cdot)^{-1}([0,+\infty))=pr_2((\{x\}\times Y)\cap f^{-1}([0,+\infty)))=\{y\in Y\ |\ f(x,y)\geq 0\}\ ,\quad pr_2(x,y)=y\ ,$$ which is a compact subset of $Y$.
Let us now endow the space $\mathcal{K}(Y)=\{K\subset Y\ |\ \varnothing\neq K\text{ compact}\}$ with the Hausdorff distance $$d_H(K,K')=\sup_{y\in Y}(|d_Y(y,K)-d_Y(y,K')|)\ ,\quad K,K'\in\mathcal{K}(Y)\ ,$$ where $$d_Y(y,Z)=\inf_{y'\in Z}d_Y(y,y')\ ,\quad \varnothing\neq Z\subset Y\ .$$
Consider the (compact) set-valued map $F:X\ni X\mapsto F(x)\subset Y$.
Question: seen as map from the metric space $(X,d_X)$ into the metric space $(\mathcal{K}(Y),d_H)$, is $F$ Lipschitz? I.e. is there $C'>0$ such that for all $x,x'\in X$ $$d_H(F(x),F(x'))\leq C'd_X(x,x')\ ?$$
No. Let $(X,d_X)$ be any compact metric space with a metric $d_Y$ for which: $d_X$ is Lipschitz continuous with respect to $d_Y$ (i.e., $d_X\leq Cd_Y$ for some $C$), but not vice-versa.
(For example, take $X=[0,1]$, $d_X$ the usual metric, and $d_Y(x,y)=|\sqrt{x}-\sqrt{y}|$. Then $d_X\leq 2d_Y$ but $d_Y$ is not Lipschitz continuous with respect to $d_X$..)
Now let $f(x,y)=-d_X(x,y)$. Then $F(x)=\left\{x\right\}$ and therefore $$d_H(F(x),F(x'))=d_Y(x,x')$$