 Is it possible to prove without Continuum Hypothesis that for every uncountable subset $S$ of $\mathbb{R}$ there is a real number $x$ that splits it into two parts of the same cardinality, i.e. $\leftS \cap (\infty,x)\right=\leftS \cap (x,\infty)\right$?
 (if the answer to the first question is no) Is this statement equivalent to Continuum Hypothesis?

$\begingroup$ Am I being very dense? What value of $x$ are you proposing to use for $\mathbb{R}^+$? $\endgroup$ – Reinstate Monica Brondahl May 12 '18 at 16:59

$\begingroup$ @Brondahl You can split $\mathbb R^+$ at any positive number, and will get 2 parts (a line segment and a ray) both of the cardinality $\mathfrak c = 2^{\aleph_0}$. $\endgroup$ – Vladimir Reshetnikov May 12 '18 at 18:28

$\begingroup$ ah, right, yes Cardinality as in Countability, as in (0,1) is indistinguishable from (0,1000) and also from (3,$\infty$). I was indeed being very dense. $\endgroup$ – Reinstate Monica Brondahl May 12 '18 at 21:32
No, the statement cannot be proven in ZFC without assuming continuum hypothesis or something similar. In fact, it is equivalent to the statement that there are finitely many cardinalities between $\aleph_0$ and $2^{\aleph_0}$, so it is strictly weaker than the continuum hypothesis.
Suppose that there were infinitely many such cardinalities, then you can let $S=\bigcup_{n=1}^\infty S_n$ where $S_n\subseteq(0,1/n)$ has cardinality $\aleph_n$ to obtain a contradition.
On the other hand, if there are only finitely many such cardinalities, then $f(x)=\vert S\cap(\infty,x)\vert$ must achieve its maximum, say $\aleph_n$ ($n > 0$). If $x_0$ is the infimum of the $x\in\mathbb{R}$ such that $f(x)=\aleph_n$ then $S\cap(x_0,\infty)$ has cardinality $\aleph_n$. Choosing $y_k\in\mathbb{R}$ decreasing to $x_0$, the cardinality of $S\cap(y_k,\infty)$ must be $\aleph_n$ for large enough $k$, otherwise $S\cap(x_0,\infty)=\bigcup_k(S\cap(y_k,\infty))$ is a countable union of sets of cardinality less than $\aleph_n$, so is of cardinality less than $\aleph_n$, giving a contradiction. So, $S\cap(\infty,y_k)$ and $S\cap(y_k,\infty)$ are both of cardinality $\aleph_n$ for $k$ large enough.
Suppose that the continuum is larger than $\aleph_\omega$. Choose a subset $S_n$ of $(n,n+1)$ of cardinality $\aleph_n$ and let $S=\cup_{n=1}^\infty S_n$. Then for each $x$, $S\cap (\infty,x)$ has cardinality smaller than $S\cap (x,\infty)$.

$\begingroup$ Could I clarify, please? The accepted answer is that the following are equivalent: (1) $2^{\aleph_{0}} < \aleph_{\omega}$; (2) every uncountable subset of reals is cut by some real into equinumerous parts. Analogous results hold for cutting uncountable subsets of the plane by lines. $\endgroup$ – Avshalom Oct 2 '14 at 12:48