According to Hausdorff distance in Wikipedia, $$ \sup_{w\in M}\rvert \inf_{x\in X}d(w,x)-\inf_{y\in Y}d(w,y)\lvert =\sup_{w\in X\cup Y}\rvert \inf_{x\in X}d(w,x)-\inf_{y\in Y}d(w,y)\lvert $$ is true. I tried to prove it and thought about a special case that $X$ and $Y$ is compact. Then, there exist $x_0\in X$ and $y_0\in Y$ such that $\inf_{x\in X}d(w,x)=d(w,x_0),\inf_{y\in Y}d(w,y)=d(w,y_0)$. I got \begin{align*} \inf_{x\in X}d(w,x)-\inf_{y\in Y}d(w,y) &=d(w,x_0)-d(w,y_0)\\ &\le d(x_0,y_0). \end{align*} I do not know whether this strategy is right. Please give me its proof or hint.
By the way, I succeeded in proving below: $$ \sup_{w\in X\cup Y}\rvert \inf_{x\in X}d(w,x)-\inf_{y\in Y}d(w,y)\rvert =\max\{\max_{w\in Y}d(X,y),\max_{w\in X}d(x,Y)\}. $$ Here, $X$ and $Y$ are compact.
It is the triangular inequality in disguise. Writing $d(w,X)=\inf_{x\in X} d(w,x)$ it suffices to show (I let you figure out why) that for every fixed $w\in M$ we have: $$ d(w,X)-d(w,Y) \leq \sup_{y\in Y} d(y,X).$$
To see this let $\epsilon>0$ and pick $y_0\in Y$ so that $d(w,y_0)\leq d(w,Y)+\epsilon$. Then \begin{align} d(w,X)-d(w,Y) &\leq \inf_{x\in X} d(w,x)-d(w,y_0)+\epsilon \\ &\leq \inf_{x\in X} d(x,y_0) + \epsilon \\ &=d(y_0,X) + \epsilon\\ &\le\sup_{y\in Y}d(y,X) + \epsilon. \end{align} And $\epsilon>0$ was arbitrary.