We have
$$d(a,C)\le d(a,b)+d(b,C)\le d(a,b)+d(B,C) $$
for all $b\in B$, so, taking $\inf_{b\in B}$, we have
$d(a,C)\le d(a,B)+d(B,C)$
which is $\le d(A,B)+d(B,C)$
then take $\sup_{a\in A}$, and similarly, we can show that $d(A,c)\le d(A,B)+d(B,C)$ as well.
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
Update: I think I have a counterexample in $\mathbb{R}^2$:
Let
$$A = \bigg(\bigg[\frac{1}{n},1\bigg]\times [0,1]\bigg) \cup ([0,1]\times [2,3])$$ $$B = ([0,1]\times [0,1]) \cup ([0,1]\times [2,3])$$ $$C = ([0,1]\times [0,1]) \cup ([0,1]\times [1,3])$$
We have that:
$$C\setminus A =\bigg(\bigg[0,\frac{1}{n}\bigg)\times [0,1]\bigg) \cup ([0,1]\times [1,2))$$ and:
$$C\setminus B =[0,1]\times [1,2)$$
Trivially we can make $d_H(A,B)$ as small as we like by increasing $n$. But for $d_H(C\setminus A,C\setminus B)$ will always be greater than or equal to $1$.