$D(A,B)$ is the maximum of the upper bound on the lower bound on the distance between two elements in the sets either way. Which is to say (very informally) the "largest" of the "shortest" distance measures between the sets.
Take, for example $A=\{0, 2\}, B=\{1,2,4\}$ and $d(x,y)=\lvert x-y\rvert$
$$\begin{align} {\sup}_{a\in A}{\inf}_{{b\in B}}~d(a,b) = ~ & \sup \big\{\inf \{d(a,b): b\in B\}: a\in A\big\} \\ =~ & \sup\big\{\inf\{d(a,1), d(a, 2), d(a,4)\}: a\in A\big\} \\ = ~ & \sup\big\{\inf \{d(0,1), d(0,2), d(0,4)\},\inf\{d(2,1), d(2,2), d(2,4)\}\big\} \\ = ~ & \sup\big\{\inf\{ 1,2,4\}, \inf\{1,0,2\}\big\} \\=~ & \sup\big\{ 1,0\big\} \\=~ & 1\\[2ex] {\sup}_{b\in B}{\inf}_{{a\in A}}~d(a,b) = ~ & \sup\big\{\inf \{d(a,b): a\in A\}: b\in B\big\} \\ = ~ & \sup\big\{\inf\{d(0,b), d(2,b)\}: b\in B\big\} \\ = ~ & \sup\big\{\inf \{d(0,1), d(2,1)\},\inf\{d(0,2), d(2,2)\},\inf\{d(0,4),d(2,4)\}\big\} \\ = ~ & \sup\big\{\inf\{ 1,1\}, \inf\{2,0\}, \inf\{4,2\}\big\} \\=~ & \sup\big\{ 1,0,2\big\} \\=~ & 2\\[2ex]D(A,B)~=~&\max\{1,2\}\\ =~& 2\end{align}$$
$D(A,B)$ will be zero if $\forall a\in A~\exists b\in B: d(a,b)=0$ and $\forall b\in B~\exists a\in A: d(a,b)=0$. Every point in either set is zero distance from some point in the other set. (Notice: This is not claiming they are identical.)
If $B_n$ goes to a point $p$, then then $ \varepsilon$-ball
$B_\varepsilon (p)$ contains $B_n$ and
$d_H(A_n,A)<\varepsilon$ for $ n\geq N$ and some $N$.
$ A_n - B_n$ contains $ A_n - B_\varepsilon (p)$ and a closed
$\varepsilon$-tubular neighborhood of $A_n - B_\varepsilon (p)$
contains $A_n,\ A_n-B_n$. Hence $ d_H(A_n - B_n,A_n - B_\varepsilon
(p)) \leq \varepsilon$.
From triangle inequality $\ast$ of $d_H$, then
\begin{align*} d_H(A_n - B_n, A) &
\leq d_H(A_n - B_n,A_n-B_\varepsilon (p)) + d_H( A_n -
B_\varepsilon (p),A_n) +d_H(A_n,A) \\&\leq 2\varepsilon + d_H(A_n,A)
\\ &
\leq 3 \varepsilon \end{align*}
We have a claim $\ast$ that $d_H$ satisfies triangle inequality :
If $d_H(X,Y)=r,\ d_H(Y,Z)=R$, then $(Y)_{r+\epsilon},\
(Y)_{R+\epsilon}$ contains $X,\ Z$ respectively.
Here $(X)_{r+\epsilon}$ contains $Y$ so that $(X)_{R+r+2\epsilon}$
contains $Z$. Similarly $(Z)_{R+r+2\epsilon}$ contains $X$ so that
$d_H(X,Z)\leq r+R$.
Best Answer
Given a sequence $(S_n)_{n\in\mathbb N}$ of sets, I assume that $$\liminf_{n\rightarrow \infty} S_n=\bigcup_{n\in\mathbb N} \bigcap_{m\ge n} S_m.$$
Let $X$ be the interval $[1/3,2/3]$ endowed with the usual metric $d$. For each $x\in X$ put $\ell(x)=x-1/6$ and $u(x)=x+1/6$. For each $n\in\mathbb N$ let $p_n=1/3$, if $n$ is odd, and $p_n=2/3$, if $n$ is even. Then for each natural $n$ we have $$A_n=\{x \in X: d(p_n, [\ell(x), u(x)])=0\}=$$ $$\{x \in [1/3,2/3]: x-1/6\le p_n\le x+1/6\}.$$ So $A_n=[1/3,1/2]$, if $n$ is odd and $A_n=[1/2,2/3]$, if $n$ is even.
Let $\delta\le 1/6$ be any positive number. Then for each natural $n$ we have $$\{x \in X: d(p_n, [\ell(x), u(x)])\le\delta\}=$$ $$\{x \in [1/3,2/3]: x-1/6-\delta\le p_n\le x+1/6+\delta\}.$$ The latter set is $[1/3,1/2+\delta]$, if $n$ is odd and $[1/2-\delta,2/3]$, if $n$ is even. Then $\liminf_{n\rightarrow \infty} \{x \in X: d(p_n, [\ell(x), u(x)])\leq \delta\}=[1/2-\delta,1/2+\delta]$, so $A=\{1/2\}$, and thus $d_H(A,A_n)=1/6$ for each natural $n$.