Let $X$ be a Polish space and denote by $\mathcal{F}(X)$ the set of all closed subsets of $X$. The Effros Borel structure on $\mathcal{F}(X)$ is the smallest $\sigma$-algebra on $\mathcal{F}(X)$ containing the family $\big\{ \{ F \in \mathcal{F}(X) \, ; \, F \cap U \neq \emptyset \} \, ; \ U \subseteq X \text{ is open} \big\}$.
Now, given a closed set $A \subseteq X$, is it necessary true that the set $\{ F \in \mathcal{F}(X) \, ; \, F \cap A \neq \emptyset \}$ is Borel (i.e. belongs to the Effros Borel structure)? I strongly hope the answer is yes (since it would help me solve some other problem), but none of my attempts to prove it was successful.
I will appreciate any help.
Unfortunately the answer is negative.
Let $H=\ell^2$ be the separable infinite dimensional Hilbert space and let $S_1\subseteq H$ be the sphere of radius $1$, that is $S_1=\{x\in H\mid \|x\|=1\}$. Note that $S_1(H)$ is closed in $H$.
As shown by Christensen (see 27.6 in Kechris Classical Descriptive Set Theory) the set $\{F\in\mathcal F(H)\mid F\cap S_1(H)\neq\varnothing\}$ is $\mathbf{\Sigma}^1_1$-complete, which implies that it cannot be Borel.
Note however that if $A$ is a compact set, then $\{F\in\mathcal F(X)\mid F\cap A\neq \varnothing\}$ is indeed Borel, because $$\{F\in\mathcal F(X)\mid F\cap A\neq \varnothing\}=\bigcap_{n\in\Bbb N}\{F\in\mathcal F(X)\mid F\cap B_{1/n}(A)\neq\varnothing\},$$ where $B_r(A)=\{x\in X\mid d(x,A)<r\}$. The left to right inclusion is obvious, while if $F$ belong to the right hand side, then easily $d(F,A)=0$, and the distance between a closed set and a compact one must be realized by some point.