Writing A as disjoint sum of sets B, C such that derived sets of A, B and C are equal. Let $A$ be a set in metric space with no limit points in $A$ ($A$ and $A'$ are disjoint). Is it possible to find such two subsets $C$ and $B$ of $A$ that intersection of $B$ and $C$ is empty and $B'=C'=A'$? By $X'$ I mean derived set of $X$, I think it is standard notation, but I decided to mention that in order to avoid possible confusion. If it is not possible in general, what are necessary and sufficient conditions for such situation or are there some related results with additional assumptions?
 A: Let $A$ be any set in any metric space; I will show that there are disjoint sets $X,Y\subseteq A$ such that $X'=Y'=A'.$ (We do not need to assume that the space is separable, nor that $A$ is disjoint from $A'.$)
Let $P$ be the set of all ordered pairs $(X,Y)$ of disjoint subsets of $A,$ partially ordered so that
$$(X_1,Y_1)\le(X_2,Y_2)\iff X_1\subseteq X_2\text{ & }Y_1\subseteq Y_2.$$
For $n\in\mathbb N$ let $P_n$ be the set of all pairs $(X,Y)\in P$ such that every $3$-element subset of $X$ or $Y$ has diameter at least $\frac1n.$
By Zorn's lemma, $P_1$ has a maximal element $(X_1,Y_1),$ which can be extended to a maximal element $(X_2,Y_2)$ of $P_2,$ and so on. Thus we can choose for each $n\in\mathbb N$ a maximal element $(X_n,Y_n)$ of $P_n$ so that
$X_1\subseteq X_2\subseteq X_3\subseteq\cdots\text{ and }Y_1\subseteq Y_2\subseteq Y_3\subseteq\cdots.$
Let $X=\bigcup_{n\in\mathbb N}X_n$ and $Y=\bigcup_{n\in\mathbb N}Y_n.$ Then $X,Y$ are disjoint subsets of $A;$ I claim that $X'=Y'=A'.$ By symmetry it will be enough to show that $X'=A'.$ Clearly $X'\subseteq A'$ since $X\subseteq A.$ I have to show that $A'\subseteq X'.$
Assume for a contradiction that $A'\not\subseteq X'.$ Choose a point $a\in A'\setminus X'$ and choose $\varepsilon\gt0$ so that $B(a;\varepsilon)\cap X\subseteq\{a\}.$ Choose $n\in\mathbb N$ so that $\frac2n\lt\varepsilon.$
Now $B(a;\frac1{2n})$ contains infinitely many elements of $A,$ but at most two elements of $Y_n.$ Therefore we can choose a point $x\in A\cap B(a;\frac1{2n})$ such that $x\notin Y_n$ and $x\ne a.$
Since $(X_n,Y_n)$ is a maximal element of $P_n,$ and since $x\notin X_n,$ it follows that $(X_n\cup\{x\},Y_n)\notin P_n.$ Thus the set $X_n\cup\{x\}$ must contain a $3$-element set $\{x,y,z\}$ of diameter less than $\frac1n.$
Now $d(y,a)\le d(y,x)+d(x,a)\lt\frac2n$ and $d(z,a)\le d(z,x)+d(x,a)\lt\frac2n,$ and so $\{y,z\}\subseteq B(a;\frac2n)\cap X_n\subseteq B(a;\varepsilon)\cap X,$ contradicting the assumption that $B(a;\varepsilon)\cap X\subseteq\{a\}.$
Corollary. For any set $A$ in any metric space, there are infinitely many pairwise disjoint sets $X_1,X_2,X_3,\dots\subseteq A$ such that $X_n'=A'$ for each $n\in\mathbb N.$
A: As an example of this phenomenon $A = \{\frac{1}{n}: n \in \mathbb{N}(=\{1,2,3,\dots\})\}, B = \{\frac{1}{2n}: n \in \mathbb{N}, C =\{\frac{1}{2n+1}: n \in \mathbb{N}\}$ all have as the derived set $\{0\}$ and $A = B \cup C, B \cap C = \emptyset$.
