$\left(*\right)$ Given an infinite set $X$ and a partition of $X$ to $X=A\cup B $ where $A\cap B=\emptyset$, then either $\left|X\right|=\left|A\right|$ or $\left|X\right|=\left|B\right|$
I'd like to prove that $\left(*\right)$ is equivalent to the axiom of choice. The direction from choice to $\left(*\right)$ is pretty easy, since choice gives $2\left|X\right|\leq\left|X^{2}\right|\leq\left|X\right|$, and so if both $\left|A\right|<\left|X\right|$ and $\left|B\right|<\left|X\right|$ we'd have$$\left|X\right|=\left|A\uplus B\right|\overbrace{\leq}^{w.l.o.g.}2\left|A\right|<2\left|X\right|\leq\left|X\right|$$
which is a contradiction. (I've also used choice for either $\left|A\right|\leq\left|B\right|$ or $\left|B\right|\leq\left|A\right|$)
In the other direction I'm stuck. The theorems I know about which are equivalent to the axiom of choice are
• The well ordering theorem
• Zorn's lemma
• Trichotomy
• Cartesian product of non empty sets is non empty
Out of these, the only one which I think I can prove from $\left(*\right)$ is trichotomy, but I was unsuccessful in doing so. A hint would be greatly appreciated.
There is a simple trick to it: Take $A$ to be any set and $B$ its Hartogs number, namely, $B$ is the least ordinal which does not inject into $A$. Now since $|X|=|A|$ or $|X|=|B|$, you can conclude that $|A|<|B|$, and therefore $A$ can be well-ordered.
A different trick would be to simply take any two infinite sets $A$ and $B$, and conclude that since $|A|\leq|X|$ and $|B|\leq|X|$, and $|X|=|A|$ or $|X|=|B|$ it has to be the case that either $|A|\leq|B|$ or $|B|\leq|A|$, so trichotomy holds.