Let $X$ be a profinite space.(i.e. $X$ is the inverse limit of finite discrete space.) Suppose $X$ is 2nd countable. Then fix any open equivalence relation $R$ on $X$. Pick $x\in X$. Then $Rx\subset X$ is a finite union of basic open sets.
$\textbf{Q:}$ Why $Rx$ equivalence class is a finite union of basic open sets?
$\textbf{Q':}$ It follows that there are only countably many open equivalence relation on $X$. What cardinality arithmetic is used here? I knew $X$ is compact and there are only finite covering required. How do I know there is a countably many open relations?
Ref. Profinite Groups Luis Ribes, Chpt 1
Q has been answered by the above comments. We know that there are only finitely many equivalence classes, each of these being the finite union of basic open sets.
$X$ has a countable base $\mathcal{B}$, therefore the set of all finite sequences in $\mathcal{B}$ is also countable and we conclude that the set $\mathcal{B}^\ast$ of all finite unions of basic open sets must be countable. Hence the set $\mathcal{B}^{\ast \ast}$ of all finite sequences in $\mathcal{B}^\ast$ is countable. The set of open equivalence relations on $X$ can be identified with a subset of $\mathcal{B}^{\ast \ast}$. This answers Q'.