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.
"The real numbers are uncountable" means that, in the set-theoretic universe where we have defined "the set of natural numbers" and "the set of real numbers", there is not a function that is a bijection between these two sets.
It means nothing more, and nothing less than that.
There are all sorts of traps, mistakes, and really subtle misunderstandings one can run into by trying to ascribe more meaning to this statement than it actually has.
You may find Skolem's paradox an interesting topic to read about, given that it involves a rigorous and precise way to see the real numbers as countable in a sense different from what is meant by "the real numbers are uncountable", and the consequent difficulties people have trying to unravel what's going on.
Best Answer
Notice that $\Bbb R \times \{0\} \subset \Bbb R \times \Bbb R$ has the same quantity of elements that $\Bbb R$¹. So if you prove that $\Bbb R$ is uncountable, you're done.
¹ Actually $|\Bbb R| = |\Bbb R^n|$ for every $n$.