What has to be proven is: If $\mathit X$ is any infinite set, then there exists two disjoint subsets $\mathit Y$ and $\mathit Z$ of $\mathit X$ such that$|\mathit X|$=$|\mathit Y|$=$|\mathit Z|$ and $\mathit X$=$\mathit Y \cup\mathit Z$.
I have used contrapositive to successfully proved the existence of A to be subset of an infinite set X such that A and its complement set A$^c$ on X to be both infinite, which is $\bf {if}$ for all subset of X, say A, A or its complement A$^c$ are finite, $\bf then$ X is finite.
It suffices to prove that there exists A such that cardinality of A and A$^c$ to be same. Since by cardinal arithmetic, |X|=|A$\cup$A$^c$|=|A|+|A$^c$|=big one in {A,A$^c$}. If I proved the existence of such an A, then it follows that |A|+|A$^c$|=|A|=|A$^c$|,which indicates |A|=|A$^c$|=|X|, then proof can be finished. However, I have no idea how to prove the existence of these two sets whose cardinality are the same. Could someone give me a hint?
So suppose that $\varphi : X \times X \mapsto X$ is a bijection. Take $a \in X$. Then $Y=\varphi[X \times \{a\}]$ and $Z=\varphi[X \times (X \setminus \{a\})]$ answer your question.