If the underlying metrics are equivalent, then so are their induced Hausdorff metrics Let $(E, d)$ be a metric space and $\mathcal C$ the set of all compact subsets of $E$. We define $d_H: \mathcal C \times \mathcal C \to \mathbb R$ by
$$
d_H (A, B) := \max \left  \{  \max_{x \in B} d(x, A),  \max_{y \in A} d (y, B) \right \}\quad \forall A,B \in \mathcal C.
$$
with $d(x, A) := \min_{y \in A} d(x, y)$. I proved that $d_H$ is indeed a metric. It is mentioned in this Wikipedia page that the topology of $\mathcal C$ depends on that of $E$, not on $d$.

I formalize it as follows.

Theorem: Let $d'$ be another metric on $E$ that is equivalent to $d$, i.e., $d, d'$ give rise to the same topology. Let $d'_H$ be the Hausdorff metric on $\mathcal C$ that is induced from $d'$. Then $d_H, d'_H$ give rise to the same topology on $\mathcal C$.

Could you elaborate on how to prove this claim?
 A: I'm following the idea from the @Alessandro Codenotti's comment.
Define:

*

*$$K_r:=\bigcup_{k\in K}B(k,r) = \{x\in X:d(x,K)<r\}.$$


*$\mathcal T$ -- topology on $E$ generated by $d$.


*for $U_1,U_2,\ldots,U_k\in\mathcal T$
$$\mathcal U(U_1,U_2,\ldots,U_k)=\left\{C\in\mathcal C:C\subset \bigcup_{i=1}^kU_i\text{ and }\forall_i\,C\cap U_i\neq\emptyset\right\}.$$


*$\mathscr U=\{\mathcal U(U_1,U_2,\ldots,U_k):k\in\Bbb N,\ U_1,U_2,\ldots,U_k\in\mathcal T\}$
We claim that $\mathscr U$ is a basis for the topology generated by $d_H$. Observe that this means that the topology generated by $d_H$ can be expressed by the topology $\mathcal T$.
Proof.

*

*Take any $U_1,U_2,\ldots,U_k\in\mathcal T$ and $C\in \mathcal U:=\mathcal U(U_1,U_2,\ldots,U_k)$. Let $c_i\in C\cap U_i$ for all $i$. We can find $r>0$ such that $B(c_i,r)\subset U_i$ and $C_r\subset \bigcup_i U_i$ (from Lemma 1.). We claim that $B_{d_H}(C,r)\subset \mathcal U$. To show it consider any $K\in \mathcal C$ such that $d_H(C,K)<r$. Then $K\subset C_r\subset \bigcup_i U_i$. Moreover for all $i$ we have $c_i\in C\subset K_r$, so $$K\cap U_i\supset K\cap B(c_i,r)\neq\emptyset.$$

*Take any $C\in \mathcal C$ and $r>0$. Since $C$ is  is totally bounded, there exist $c_1,c_2\ldots,c_k\in C$ such that $$C\subset \bigcup_i U_i,\quad \text{ where } U_i=B(c_i,r/3).\tag{1}$$ We claim that $\mathcal U:=\mathcal U(U_1,\ldots,U_k)\in B_{d_H}(C,r)$. To show it consider any $K\in \mathcal U$. Then $$K\subset \bigcup_iB(c_i,r/3)\subset C_{r/3}.\tag{2}$$
Since $K\cap B(c_i,r/3)\neq \emptyset$, we have $c_i\in K_{r/3}$ and therefore $U_i=B(c_i,r/3)\subset K_{2r/3}$. From (1) we thus get $$C\subset K_{2r/3}.\tag{3}$$
From (2) and (3) we know that $d_H(C,K)\leq 2r/3<r$.


We used the following two lemmata.
Lemma.  If $K$ is compact, $U$ is open and $K\subset U$ then $K_r\subset U$ for some $r>0$.
Sketch of the proof. The function $d(\cdot,E\setminus U)\colon K\to (0,\infty)$ attains it's minimum $r>0$. $\ \square$
Lemma. We have $d_H(C,K)=\inf\{r>0:C\subset K_r,\ K\subset C_r\}$. $\ \square$
Corollary.

*

*if $d_H(C,K)<r$ then $C\subset K_r$ and $K\subset C_r$

*if $C\subset K_r$ and $K\subset C_r$ then $d_H(C,K)\leq r$.

A: Let $A, A_n \in \mathcal C$ such that $d_H (A, A_n) \to 0$. By this result, we get
$$
A = \{ \lim x_n \mid  (x_n) \in (A_n)\text{ convergent in } d\}.
$$
We need to prove that $d'_H (A, A_n) \to 0$. Assume the contrary that $d'_H (A, A_n) \not\to 0$. Then there is $\varepsilon > 0$ and a subsequence $\varphi$ of $\mathbb N$ such that $d'_H (A, A_{\varphi (n)}) > \varepsilon$ for all $n$. By definition of $d'_H$, we have $2$ cases.

*

*There exists a subsequence $\psi$ of $\varphi$ and a sequence $(\ell_{\psi (n)}) \subset A$ such that $d' (\ell_{\psi (n)}, A_{\psi (n)}) > \varepsilon$ for all $n$.

Because $A$ is compact, there is a subsequence $\lambda$ of $\psi$ and $\ell \in A$ such that $\ell_{\lambda (n)} \xrightarrow{n \to \infty} \ell$ in $d'$.  Then
$$
d' (\ell, A_{\lambda (n)}) \ge d' (\ell_{\lambda (n)}, A_{\lambda (n)})- d' (\ell, \ell_{\lambda (n)}) > \varepsilon/2
$$
for $n$ large enough. This implies if $(x_{\lambda (n)}) \in (A_{\lambda (n)})$ then $x_{\lambda (n)} \not \to \ell$ in $d'$. Because $d,d'$ are equivalent, if $(x_{\lambda (n)}) \in (A_{\lambda (n)})$ then $x_{\lambda (n)} \not \to \ell$ in $d$. This implies $\ell \notin A$, which is a contradiction.


*There exists a subsequence $\psi$ of $\varphi$ and a sequence $(a_{\psi (n)}) \in (A_{\psi (n)})$ such that $d' (a_{\psi (n)}, A) > \varepsilon$ for all $n$.

Because $A_n \to A$ in $d_H$, there is a subsequence $\lambda$ of $\psi$ and sequences $(a_{\lambda(n)}) \in (A_{\lambda (n)})$ and $(\ell_{\lambda(n)}) \subset A$ such that $d(a_{\lambda(n)}, \ell_{\lambda(n)}) \to 0$. Because $A$ is compact, there is a subsequence $\eta$ of $\lambda$ and $\ell^* \in A$ such that $\ell_{\eta (n)} \xrightarrow{n \to \infty} \ell^*$ in $d$. It follows that $a_{\eta (n)} \xrightarrow{n \to \infty} \ell^*$ in $d$.
It follows from $d' (a_{\psi (n)}, A) > \varepsilon$ for all $n$ that
$$
d' (a_{\eta (n)}, \ell) > \varepsilon \quad \forall n \in \mathbb N, \forall \ell \in A.
$$
Hence $a_{\eta (n)} \not \to \ell$ in $d'$ for all $\ell \in A$. Because $d,d'$ are equivalent, $a_{\eta (n)} \not \to \ell$ in $d$ for all $\ell \in A$. In particular, $a_{\eta (n)} \not \to \ell^*$ in $d$. This is a contradiction.
