Let $S$ be a set of cardinality $c$. Let $x$ and $y$ be two distinct elements in $S$.

Question

Let $S$ be a set of cardinality $c$. Let $x$ and $y$ be two distinct elements in $S$. How to prove that there exist two disjoint subsets $X$ and $Y$ of $S$, each of cardinality $c$, such that $x \in X$ and $y \in Y$?

Answer 1

Take $a \neq b$. Since $ (\mathbb{R} \times \{a \}) \cup (\mathbb{R} \cup \{b\})$ has cardinality $\mathfrak{c}$ there exists a bijection $$ f: (\mathbb{R} \times \{a \}) \cup (\mathbb{R} \cup \{b\}) \to S$$ Let $A=f \left[ \mathbb{R} \times \{a \} \right]$ and $B=f \left[ \mathbb{R} \times \{b \} \right]$, so that $S=A \cup B$, with $A$ and $B$ both having cardinality $\mathfrak{c}$. All it takes now is applying some bijection $S \to S$ to get your two sets.

Answer 2

Since $S$ has cardinality $\mathfrak{c}$ there is a bijection $ f: S \to \mathbb{R}$. Moreover $x\not=y$ implies that $f(x)\not=f(y)$. We may assume that $f(x)<f(y)$. Let $a=\frac{f(x)+f(y)}{2}$ then $f(x)<a<f(y)$. Take $X:=f^{-1}((-\infty,a))$ and $Y:=f^{-1}((a,+\infty))$. Hence $x\in X$, $y\in Y$, $X\cap Y=\emptyset$ and $|X|=|Y|= \mathfrak{c}$.

P.S. $x\to \ln|x-a|$ is a bijection from $(a,+\infty)$ to $\mathbb{R}$, and a bijection from $(-\infty, a)$ to $\mathbb{R}$.