Let $\newcommand{\om}{[\omega]^\omega}\om$ denote the collection of infinite subsets of the set of non-negative integers $\omega$. We say $A\subseteq\om$ is separating if for all $n,m\in\omega$ with $n\neq m$ there are $a,b\in A$ such that $n\in a, m\in b$ and $a\cap b=\varnothing$.
Question. Is there $A\subseteq \om$ such that $A$ is separating, but for all $a\in A$ the set $A\setminus\{a\}$ is no longer separating?
It will suffice to find a minimal separating set $A\subseteq[\mathbb Z]^\omega$. For this let $A=\bigcup_{n\in\mathbb Z}\{a_n,b_n\}$ where $a_n=\{x\in\mathbb Z:x\le n\}$ and $b_n=\{x\in\mathbb Z:x\gt n\}$. Note that $a_n$ and $b_n$ are the only sets in $A$ separating $n$ and $n+1$.
More generally, to get a minimal separating subset of $[\kappa]^\kappa$ where $\kappa$ is an infinite cardinal, take a totally ordered set $S$ of order type $\kappa^*+\kappa$ and let $$A=\bigcup\{\{a_s,b_s\}:s\in S,\ s\text{ has an immediate successor}\}$$ where $a_s=\{x\in S:x\le s\}$ and $b_s=\{x\in S:x\gt s\}$.
That $A$ is a separating subset of $[S]^\kappa$ follows from the fact that, if $s,t\in S$, $s\lt t$, then any element between $s$ and $t$ has an immediate successor or an immediate predecessor.
Let's say that a set $B\subseteq[S]^\kappa$ is weakly separating if, whenever $s,t\in S$ and $s\ne t$, there is a set in $B$ containg $s$ but not $t$. No proper subset of $A$ is even weakly separating since, if $t$ is an immediate successor of $s$, then $a_s$ is the only set in $A$ containing $s$ but not $t$, while $b_s$ is the only set in $A$ containing $t$ but not $s$.
Yes, consider the sets $$ a_{i,d}:=\big\{m\in \omega \;\big|\; \text{the $i$-th binary digit of $m$ is $d$}\big\} $$ and the collection $A=\{a_{i,d}\mid i\in\omega,\; d\in\{0,1\}\}$.
