Does any uncountable set contain two disjoint uncountable sets? Is it true that for any uncountable $S$, there exits two uncountable subsets $S_1,S_2 \subseteq S$ with $S_1 \cap S_2 = \emptyset$? 
I can find no counter example, but no proof either.
I am aware of similar questions regarding the partition of uncountable subsets of the real line, and there are many examples of disjoints uncountable subsets of $\mathbb{R}$, say $R_1,R_2$. 
It is not clear to me however whether this extends to any uncountable set. For uncountable sets $S$ with $S \succeq_{card} \mathbb{R}$, there exists an injection $f : \mathbb{R} \rightarrow S$ and choosing $S_1,S_2$ with $S_i = \{s\in S~|~ s = f(r_i) $ for some $ r_i \in R_i\}$, works. 
But from what I understood of the continuum hypothesis, there might exist an uncountable sets $H$ with $\mathbb{R} \succeq_{card} H$. So using the above argument, one cannot infer that the property holds for every uncountable sets from observing that it holds for $\mathbb{R}$. Is this right? Are there other ways to prove or disprove the assertion?
 A: Under the axiom of choice, for any two infinite cardinal numbers, their sum, which corresponds to their disjoint union, is simply the maximum of the two numbers. This is a standard fact of cardinal arithmetic you can find in every set theory book.
So if $S$ is uncountable with cardinal $\alpha$, we have $\alpha+\alpha=\alpha$. So their exists two disjoint copies $S_1, S_2$ of $S$ such that their union has the same cardinality as $S$. So there is a bijection $f:S_1\cup S_2\to S$. So $f(S_1)$ and $f(S_2)$ are two disjoint uncountable subsets of $S$.

Here is an actual method to "construct" a bijection from $S$ to $S\times\{0,1\}$. Let $\mathcal{F}$ be the family of all bijections of some subset $T$ of $S$ to $T\times\{0,1\}$. Such a bijection always exists when $T$ is a countably infinite subset. Order these functions by set inclusion on their graph. Then the conditions of Zorn's Lemma are satisfied, and their exists a maximal such function of the form $f:T\to T\times\{0,1\}$. If $S\backslash T$ would be infinite, it would contain a countable subset and then we could extend $f$ to a larger such function in contradiction to it being maximal. So $S\backslash T$ must be finite. So there is a bijection $g:\{0,1,\ldots,n-1\}\to S\backslash T\times\{0,1\}$. Let $C=\{x_0,x_1,\ldots\}\subseteq S$ be countably infinite. You can construct now a new bijection $f':S\to S\times\{0,1\}$ by 
$$f'(x) = \begin{cases} g(m) &\mbox{if } x_m\in\{x_0,\ldots,x_{n-1}\}\subseteq C\\ 
f(x_{m-n}) & \mbox{if } x_m\in C\text{ and }m\geq n.\\
f(x) &\mbox{otherwise.}\end{cases} $$
The proof is essentially from Halmos little set theory book.
