Let $(X,\tau)$ be a topological space (eventually, a polish space) and $\mathcal{F}$ the collection of all closed sets of $X$. Given $\mathcal{U}\subseteq \tau$ finite, define $$ \mathcal{V}(\mathcal{U})= \begin{cases} \left\{F\in\mathcal{F}(X):F\subseteq\bigcup\mathcal{U}\wedge\forall U\in\mathcal{U}(F\cap U\neq\emptyset)\right\} &,\text{ if } \mathcal{U\neq\emptyset}; \text{ and }\\ \{\emptyset\}&,\text{ if }\mathcal{U}=\emptyset \end{cases} $$ Then $\mathcal{B}=\{\mathcal{V}(\mathcal{U}):\mathcal{U}\subseteq\tau\text{ is finite }\}$ is a base for some topology of $\mathcal{F}(X)$, called the Vietoris topology. Given a open set $U$ of $X$, let $$\mathcal{V}_0(U) = \{F\in\mathcal{F}(X):F\cap U\neq\emptyset\}$$ and $\mathcal{E}$ be the $\sigma$-algebra on $\mathcal{F}(X)$ generated by the collection $\mathcal{B}_0=\{\mathcal{V}_0(U):U\in\tau\}$. Then the mensurable space $(\mathcal{F}(X),\mathcal{E})$ is the Effros's Borel structure of $X$.
Now, let $Y$ be a topological space and $f:\mathcal{F}(X)\to Y$ a continuous function (with respect to the Vietoris topology). Is this function Borel measurable with respect to the Effros's Borel structure? I do think the answer seems to be yes, as
$$\mathcal{P}=\big\{\{F\in\mathcal{F}(X):F\cap U\neq\emptyset\}:U\in\tau\big\}\cup \big\{\{F\in\mathcal{F}(X):F\subseteq U\}:U\in\tau\big\}$$ is a subbase for the Vietoris topology. But, by a commentary of Kechris's Classical Descriptive Set Theory, I know that sets of form $\{\{F\in\mathcal{F}(X):F\subseteq U\}$, with $U\in\tau$, need not to be Borel in the Effros's structure.