The Problem: Let $X$ be metrizable, let $K(X)$ be the space of all compact subsets of $X$ with the topology induced by the Hausdorff metric $d_H$. Show that the set $K_f(X)=\{A\in K(X): K\,\text{is finite}\}$ is of the form $\underset{n\in\mathbb{N}}{\cup}F_n$, where $F_n$ are closed sets.
Source: Classical Descriptive Set Theory by Alexander S. Kechris.
My Attempt: Let $\mathcal{A}_n=\{A\in K(X): |A|\leq n\}$. It suffices to show that each $\mathcal{A}_n$ is closed. To that end, fix $n\in\mathbb{N}$, and let $(A_i)$ be a sequence in $\mathcal{A}_n$ such that $(A_i)\to A$ for some $A\in K(X)$. We show that $A\in\mathcal{A}_n$. Note that $(A_i)\to A\iff d_H(A, A_i)\to0\iff\forall\epsilon>0\exists N\in\mathbb{N}\;\text{such that}\;i\geq N$ implies $\underset{x\in A}{\max}d(x, A_i)<\epsilon$ and $\underset{y\in A_i}{\max}d(y, A)<\epsilon$. Recall that $d(x, A_i)=\underset{y\in A_i}{\inf}d(x, y)$ and similarly $d(y, A)=\underset{x\in A}{\inf}d(y, x)$. Suppose, BWOC, that $|A|\geq n+1$. Screeching halt.
Any help would be greatly appreciated.
We argue by contradiction. Let $(K_i)_{i\in \mathbb{N}} \subseteq \mathcal{K}_n$ that converges to $S\in K(X)$. Assume that $\vert S \vert\geq n+1$. Pick $n+1$ (pairwise different) points $(x_j)_{j=1}^{n+1}$ in $S$.
Furthermore, let $\ell = \min\{ d(x_i, x_j) \ : \ i,j\in \{1, \dots, n+1\}, i\neq j \}$ be the minimal distance between the points $(x_j)_{j=1}^{n+1}$ and choose $\varepsilon = \ell/4$.
Let $N_\varepsilon\in \mathbb{N}$, such that $d_H(S,K_j)<\varepsilon$ for all $j\geq N_\varepsilon$. As $d_H(x_i,K_j)\leq d(S,K_j) < \varepsilon$, we get that $\vert B(x_i,\varepsilon) \vert \geq 1$. On the other hand, two such balls cannot intersect, i.e. $B(x_j,\varepsilon) \cap B(x_m, \varepsilon) = \emptyset$ for all $j,m\in \{1, \dots, n+1\}$ with $j\neq m$. Indeed, assume that $x\in B(x_j,\varepsilon) \cap B(x_m, \varepsilon)$, then we have
\begin{align*} \ell < d(x_j, x_m) \leq d(x_j,x) + d(x,x_m) < \varepsilon + \varepsilon = \ell/2, \end{align*} which is a contradiction.
However, we have now $n+1$ balls that need to contain at least one point from $K_j$ and no point in $K_j$ can be in two ball simultaneously, thus $K_j$ contains at least $n+1$ points, which contradicts the assumption that $K_j\in \mathcal{K}_n$.
This proves that $S\in \mathcal{K}_n$ and hence $\mathcal{K}_n$ is a closed subset of $K(X)$. Therefore, we can write
$$ K_f(X) = \bigcup_{n\in \mathbb{N}} \mathcal{K}_n, $$
i.e. $K_f(X)$ is a countable union of closed sets in $K(X)$.