TITLE: Is there any infinite set which has no disjoint infinite subsets?
QUESTION [9 upvotes]: Is there any infinite set which has no disjoint infinite subsets?
I claim that the answer is NO!  

Proof:
  Let $S$ be any infinite set. Take a countable subset $A$ of $S.$  Then we can find a bijection $f:A\to\mathbb{N}.$ Define $B=\{b\in A:f(b)\, \text{is even}\}$ and $C=\{c\in A:f(c)\,\text{is odd}\}.$
  Then clearly $B$ and $C$ are disjoint infinite subsets of $S.$
  Hence we are done. 

Am I correct?
If this is correct, this proof is depend on the fact that "Every infinite set has a countable subset."
Can we prove the same result without using above fact?

REPLY [15 votes]: Assuming the axiom of choice, yes. Every infinite set has a countably infinite subset. Since we can split a countable set to two disjoint infinite sets, we can split the original set into two disjoint infinite sets. 
However, without the axiom of choice it is consistent that there is an infinite set which cannot be split into two disjoint infinite sets. Such a set is called amorphous.