Since the answer to my previous question (see Effros Borel structure ) turned out to be negative, I don't know how to solve the following problem:
Let $X$ be a Polish space and let $\mathcal{F}(X)$ be the space of all closed subsets of $X$ equipped with the Effros Borel structure. Is it necessary true that $\mathcal{K}(X)$, the set of all compact subsets of $X$, is Borel as a subset of $\mathcal{F}(X)$?
Trivially, the answer is positive if $X$ is compact. However, the case I am interested in is $X = \mathbb{U}$, where $\mathbb{U}$ stands for the universal Urysohn space (which is not compact).
Yes, $\mathcal K(X)$ is indeed Borel in $\mathcal F(X)$.
In fact when $X$ is locally compact and $\mathcal F(X)$ has the Fell topology, $\mathcal K(X)$ is $\mathbf{\Pi}^0_3$ at worst. To see that recall that a subspace of a metrizable space is compact iff it is complete and totally bounded. Since we are only dealing with closed sets we need to worry only about being totally bounded. Let $D\subseteq X$ be a dense countable set, then $F\in\mathcal F(X)$ is totally bounded iff for all $n\in\Bbb N$ there is a finite $A\subseteq D$ such that $F\subseteq\overline{\bigcup_{a\in A}B_{1/n}(a)}$, which means that we can write:
$$\mathcal K(X)=\bigcap_{n\in\Bbb N}\bigcup_{A\in[D]^{<\omega}}\left\{F\in\mathcal F(X)\mid F\subseteq \overline{\bigcup_{a\in A}B_{1/n}(a)}\right\}.$$
(I'm taking closure of the union of balls even though it wouldn't be needed for total boundedness to make sure that the set is in the Borel $\sigma$-algebra, being the complement of a generating set)