Let $X$ be the collections of all non-empty compact subsets of $\mathbb{R}^2$, which has the Euclidean metric. Let $(X,d)$ be a metric space, where $d$ is the Hasudorff metric.
Is $X$ separable?
Furthermore, let $Y$ be the collections of all finite subsets of $X$. Let $(Y,d)$ be a metric space, where $d$ is the Hasudorff metric.
Is $Y$ separable?
Let us suppose, more generally, that $(M, \rho)$ is a separable metric space, with countable and dense subspace $N$. Let $X$ be the set of non-empty compact subsets of $M$, $Y \subseteq X$ be the set of non-empty finite subsets of $M$, and $Z \subseteq Y$ be the non-empty finite subsets of $N$ (note that $Z$ is countable). Let $d$ be the Hausdorff metric. I claim that $Z$ is dense in $Y$, $Y$ is dense in $X$, hence $Z$ is dense in $X$, making both $X$ and $Y$ separable.
Fix $A \in X$ and $\varepsilon > 0$. Since $A$ is compact, it is totally bounded, and so we may find an $\varepsilon$-net, i.e. a finite set $a_1, \ldots, a_n \in A$ such that $$A \subseteq \bigcup_{i=1}^n B(a_i; \varepsilon). \tag{1}$$ In particular, the finite set $A' = \{a_1, \ldots, a_n\}$ is a subset of $A$, hence $$\sup_{x \in A'} \inf_{y \in A} \rho(x, y) = 0.$$ Moreover, $(1)$ implies that \begin{align*} \sup_{x \in A} \inf_{y \in A'} \rho(x, y) &\le \sup_{x \in \bigcup_{i=1}^nB(a_i; \varepsilon)} \inf_{j \in \{1, \ldots, n\}} \rho(x, a_j) \\ &= \max_{i \in \{1, \ldots, n\}} \sup_{x \in B(a_i; \varepsilon)} \inf_{j \in \{1, \ldots, n\}} \rho(x, a_j) \\ &\le \max_{i \in \{1, \ldots, n\}} \sup_{x \in B(a_i; \varepsilon)} \rho(x, a_i) \\ &= \varepsilon. \end{align*} Thus, $d(A', A) \le \varepsilon$, hence $Y$ is dense in $X$.
To show $Z$ is dense in $Y$, we can show a quick lemma:
Suppose $A = \{x_1, \ldots, x_n\}$ and $B = \{y_1, \ldots, y_n\}$ are subsets of a metric space $(M, \rho)$, and $d$ is the Hausdorff metric. Then $$d(A, B) \le \max_i \rho(x_i, y_i).$$
To prove this, we have \begin{align*} \sup_{x \in A} \inf_{y \in B} \rho(x, y) &= \max_{i=1, \ldots n} \min_{j = 1, \ldots, n} \rho(x_i, y_j) \\ &\le \max_{i=1, \ldots n} \rho(x_i, y_i). \end{align*} A symmetric argument shows that $$\sup_{y \in B} \inf_{x \in A} \rho(x, y) \le \max_{i=1, \ldots n} \rho(x_i, y_i)$$ as well. $\square$
Now, suppose $B = \{b_1, \ldots, b_m\} \in Y$ and $\varepsilon > 0$. As $N$ is dense in $M$, for $i = 1, \ldots, m$, there exists some $c_i \in N$ such that $\rho(c_i, b_i) < \varepsilon$. If we let $C = \{c_1, \ldots, c_m\}$, by the previous lemma, we have $d(B, C) < \varepsilon$.
Thus, $Z$ is countable and dense in $Y$, and hence in $X$. Both $X$ and $Y$ are separable.
