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?

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.