Proving that an uncountable set has an uncountable subset whose complement is uncountable. How does one prove that an uncountable set has an uncountable subset whose complement is uncountable. I know it needs the axiom of choice but I've never worked with it, so I can't figure out how to use. Here is my attempt (which seems wrong from the start): 
Let $X$ be an uncountable set, write $X$ as a disjoint uncountable union of the sets $\{x_{i_1},x_{i_2}\}$ i.e $X=\bigcup_{i\in I}\{x_{i_1},x_{i_2}\}$ where $I$ is an uncountable index (I'm pretty sure writing $X$ like this can't always be done), using the axiom of choice on the collection $\{x_{i_1},x_{i_2}\}$ we get an uncountable set which say is all the ${x_{i_1}}$ then the remaining ${x_{i_2}}$ are uncountable.
Anyway how is it done, properly?
I know the question has been asked in some form here but the answers are beyond my knowledge. 
 A: Your idea is generally correct.
Using the axiom of choice, $|X|=|X|+|X|$, so there is a bijection between $X$ and $X\times\{0,1\}$. Clearly the latter can be partitioned into two uncountable sets, $X\times\{0\}$ and $X\times\{1\}$.
Therefore $X$ can be partitioned to two uncountable disjoint sets.

Indeed you need the axiom of choice to even have that every infinite set can be written as a disjoint union of two infinite sets, let alone uncountable ones.
A: Sorry for the necropost; I came across this and wanted to share a proof using only Zorn's lemma (i.e. an "elementary" proof).
Edit: "elementary" might not be the best word to use here. Perhaps "easy" is better. See comments.

Let
$
P=\left\{ A \subset X\times X\colon\phi(A)\right\} 
$
where $\phi$ is the proposition given by:
$$
\phi(A)=\forall(x,y)\in X\times X\colon(x,y)\in A\implies\psi(A,x,y)
$$
and
$$
\psi(A,x,y)=\forall(w,z)\in A\colon\left(x=w\iff y=z\right)\wedge x\neq z\wedge y\neq w.
$$

Example: If $X=\mathbb{N}$, the set $A=\{(1,2),(3,4)\}$ would be in $P$, but the set
  $A^{\prime}=\{(1,2),(3,1)\}$ would not (the number $1$ appears
  twice).

Define a partial order on $P$ by inclusion $\subset$. Trivially, every chain in $P$ has an upper bound given by the union of the elements in that chain. By Zorn's lemma, $P$ has a maximal element, say
$$
A^{\star}=\{(x_{\alpha},y_{\alpha})\}.
$$
Let $X_{1}=\{x_{\alpha}\}$ and $X_{2}=\{y_{\alpha}\}$. Then $X_{1}$ and $X_{2}$ are disjoint by construction and necessarily uncountable (otherwise $X$ is not uncountable). Let 
$
Z= X_{1}\sqcup X_{2}
$
and note that $X\setminus Z$ has at most one element, for otherwise we contradict the maximality of $A^{\star}$. Lastly, let 
$$
X_{1}^{\prime}= X_{1}\sqcup(X\setminus Z),
$$
so that $X= X_{1}^{\prime}\sqcup X_{2}$, as desired.
