This is, actually, part of my research work. I'm finding it hard to understand how it came about since I have to explain it later.
Let $X=\mathbb{R}$ be with the usual metric and $K(X)$ be the family of nonempty compact subsets of $X.$ Define multivalued map, $ T:X\rightarrow K(X)$ by $$ \begin{eqnarray} Tx=\begin{cases} \left[-4x,-\dfrac{7}{2}x \right],&x\in \left[0,\infty \right),\\ \left[-\dfrac{7}{2}x ,-4x\right],&x\in \left(-\infty, 0\right]. \end{cases} \end{eqnarray} $$ Suppose $x,y\in \left(0,\infty \right)\cup \left(-\infty,0 \right),$ then \begin{align} H(Tx,Ty)&=\max\left\lbrace \sup\limits_{a\in Tx}dist(a,Ty),\sup\limits_{b\in Ty}dist(Tx,b) \right\rbrace \\&=\max\left\lbrace \sup\limits_{a\in Tx}\left( \inf\limits_{b\in Ty} d(a,b)\right) ,\sup\limits_{b\in Ty}\left( \inf\limits_{a\in Tx} d(a,b)\right)\right\rbrace\\&=\max\left\lbrace \sup\limits_{a\in Tx}\left( \inf\limits_{b\in Ty} |a-b|\right) ,\sup\limits_{b\in Ty}\left( \inf\limits_{a\in Tx} |a-b|\right) \right\rbrace\\&=\max\left\lbrace \sup\limits_{a\in Tx} \left\lbrace |a+4y|\right\rbrace ,\sup\limits_{b\in Ty}\left\lbrace \left| -4x-b\right| \right\rbrace \right\rbrace\\&=\max\left\lbrace 4\left| x-y\right|,4\left| x-y\right| \right\rbrace \\&=4\left| x-y\right| \end{align}
I think I got it all wrong when calculating $$\inf\limits_{b\in Ty} |a-b|,\inf\limits_{a\in Tx} |a-b|,$$ for fixed $a$ and $b$, respectively.
Question: Please, how do I get it? I mean $\inf\limits_{b\in Ty} |a-b|$ for a fixed $a$ in $Tx.$
We can assume without loss of generality that $0<y<x$. With this assumption we know that $$-4x<-4y \ \ \mbox{ and } \ -\frac{7}{2}x <-\frac{7}{2}y.$$ Let $a \in Tx$ so that $$-4x \leq a \leq -\frac{7}{2}x.$$ If $-\frac{7}{2}x \leq -4y$, then we have that the closest point from $Ty$ to $a$ is $-4y$ because in this case $$a \leq -4y \leq b$$ for all $b \in Ty$. Hence $d(a,Ty)=|a-(-4y)|=|a+4y|$ provided $-\frac{7}{2}x \leq -4y$.
On the other hand, if $-4y<-\frac{7}{2}x$, then we have that $$-4x<-4y < -\frac{7}{2}x<-\frac{7}{2}y.$$ From this it follows that $d(a,Ty)=0$ if $-4y \leq a \leq -\frac{7}{2}x$ because in that case $a \in Ty$. If $-4x\leq a <-4y$ then $d(a,Ty)=|a-(-4y)|$.
Summing up, $d(a, Ty)=0$ if $a \in Tx \cap Ty$, and $d(a,Ty)=|a+4y|$ if $a \in Tx$ but $a \notin Ty$.