Let $X$ be a Polish space and $\mathcal{F}(X)$ the set of closed subsets of $X$ endowed with the Effros Borel structure, generated by sets of the form $\{F\in \mathcal{F}(X):F\cap U\neq \emptyset\}$, where $U$ ranges over the open subsets of $X$.
Question: Is the set $\mathcal{C}(X)$ of all clopen subsets of $X$ a Borel subset of $\mathcal{F}(X)$? I am particularly interested in the case when $X=\mathbb{N}^\mathbb{N}$ (which is, notably, not locally compact).
One attempt to show this is to note that a closed set $F$ is clopen if and only if there is another closed set $G$ such that $F\cap G=\emptyset$ and $F\cup G=X$, but the intersection operation is not Borel on $\mathcal{F}(X)$ unless $X$ is locally compact, and even then, this appears at best a $\mathbf{\Sigma}^1_1$, or analytic, characterization. Another thought is to use that the set $\mathcal{O}(X)$ of open subsets of $X$ can be endowed with a Borel structure via complementation... but I don't see how that helps (the intersection of Borel spaces need not be Borel if their structures don't cohere in some way).
Edit: If $X$ is totally disconnected, we can fix a sequence $(U_n)$ of basic clopen subsets, an element of $\mathcal{F}(X)^\mathbb{N}$, and say that $F\in\mathcal{F}(X)$ is clopen if and only if for every $x\in F$, there is an $n$ such that $x\in U_n$ and $U_n\subseteq F$. The subset relation is Borel on $\mathcal{F}(X)$ (see section 12.C of Kechris, Classical Descriptive Set Theory), so this is a $\mathbf{\Pi}^1_1$ description of being clopen. Together with what was written above, this shows that "clopen" is a Borel property in $\mathcal{F}(X)$, provided $X$ is totally disconnected and locally compact, but still leaves out the case $X=\mathbb{N}^\mathbb{N}$, in which I am most interested.