Consider the following result by Burgess:
Theorem: Let $E$ be an analytic equivalence relation on a Polish space $X$. Then either $|X/E|\leq\aleph_1$ or there is a perfect set of pairwise $E$ inequivalent elements in $X$.
I want to prove it using the following result by Silver:
Theorem: If $E$ is a coanalytic equivalence relation on a Polish space $X$, then either $|X/E|\leq\aleph_0$ or there is a perfect set of pairwise $E$ inequivalent elements.
I can show that every analytic equivalence relation $E$ on a Polish space $X$ is of the form $$ E=\bigcap_{\alpha<\omega_1}B_{\alpha}, $$
where $B_{\alpha}$ is a Borel equivalence relation on $X$. My attempt is to show that if there are $\aleph_2$ many $E$ equivalence classes, then $B_{\alpha}$ must have uncountably many equivalence classes, for some $\alpha<\omega_1$, and then use Silver's result.
Is it possible construct a binary tree of equivalence classes in the following way:
Let $(\alpha_n:n<\omega)$ be a strictly increasing sequence in $\omega_1$ and set $\gamma:=sup\{\alpha_n:n<\omega\}$.
For every $s\in2^{<\omega}$ with $k:=|s|$, choose a $B_{\alpha_k}$ equivalence class $E_s$ such that $E_t\subseteq E_s$, for $s\subseteq t$, and if neither $s\subseteq t$ nor $t\subseteq s$, then $E_s\cap E_t=\emptyset$.
The difficulty then is to show that for every $f\in2^{\omega}$, $\bigcap_{n<\omega}E_{f\upharpoonright n}\neq\emptyset$, which would yield uncountably many $B_{\gamma}$ classes.
Any help, including other approaches, is appreciated.
This doesn't work, because even countable intersections of Borel equivalence relations with countably many equivalence classes each can have a perfect set of inequivalent elements. For example, let $B_n$ be such that $x,y\in\mathbb{R}$ are equivalent if they are in the same interval of the form $[\frac{k}{n},\frac{k+1}{n})$ for some integer $k$. Then $\cap_{n\in\mathbb{Z}_+}B_n$ is equality on the reals.