Let $X$ be a Tychonoff space, and let $K(X)$ be the space of nonempty compact subsets of $X$, equipped with the Vietoris topology. For $n \in \mathbb{N}$, define $ F_n(X) := \{ A \in K(X) : |A| \leq n \}. $ It is stated (without proof) in Cardinal Functions on Hyperspaces and Function Spaces by Ibula Ntantu that $F_n(X)$ is closed in $K(X)$, but I haven’t been able to think of a clear justification.
Any standard reference or direct proof would be appreciated.
I will use the standard basis of $K(X)$ given by $$O(U_1,\ldots,U_m)=\{A\in K(X)\mid A\subseteq\bigcup_{i=1}^m U_i\text{ and }A\cap U_i\neq\varnothing\text{ for every }i\},$$ as $m$ ranges over $\mathbb N$ and $U_i$ ranges over open subsets of $X$.
Let's show that $K(X)\setminus F_n(X)$ is open. Let $A\in K(X)\setminus F_n(X)$, that is let $A$ be a compact subset of $X$ containing at least $n+1$ points $a_1,\ldots a_{n+1}$. Using regularity of $X$ we can find pairwise disjoint open subsets of $X$, $U_1,\ldots, U_{n+1}$ such that $a_i\in U_i$ for every $i$. Then $V=O(U_1,\ldots,U_{n+1},X)$ is a neighbourhood of $A$ in $K(X)$, and it is disjoint from $F_n(X)$, since every element of $V$ meets $n+1$ pairwise disjoint open sets. This shows that $K(X)\setminus F_n(X)$ is open, or in other words that $F_n(X)$ is closed.