Can every uncountable subset of $\mathbb R$ be split up this way? For me this question is like a fish that anytime when I (seem to) catch it, manages to slip out of my hands again.

If $U$ is an uncountable subset of $\mathbb R$ then can it be shown that some $x\in\mathbb R$ exists such that $U\cap(-\infty,x)$ and $U\cap(x,\infty)$ are both uncountable?

Thanks in advance.
 A: Let $a$ be the supremum of all $x$ such that $U\cap (-\infty,x)$ is countable ($a=-\infty$ if there are none). Let $b$ be the infimum of all $x$ such that $U\cap (x,\infty)$ is countable, and $\infty$ if there are none. Now $a=b$ would imply that $U$ is countable so we must have 
$a<b$ and any $a<x<b$ will satisfy your demands. 
A: Since $U$ is uncountable (and $\mathbb{R}$ is second countable), all but countably many points of $U$ are condensation points of $U$. If $x < y$ are two condensation points of $U$, then $\frac{x+y}{2}$ is as desired.
A: I could be off here, but the following comes to mind:
If $U$ has no neither first nor last item, for instance if it's set $\mathbb{Z}$, then any splitting the subset $U$ to $U_1\in\langle-\infty,x\rangle\cap U$ and $U_2\in\langle x,\infty\rangle$, both $U_1$ and $U_2$ are uncountable.
However, if the set has a first or a last item, for example set $\mathbb{N}$, then $U_1$ xor $U_2$ would be countable.
It would be impossible to prove that both subsets are uncountable without further specifying that $U$ doesn't have first nor last item. And if it so, any $x\in\langle-\infty,\infty\rangle$ would satisfy the criteria because $U\in\langle-\infty,\infty\rangle$ and $x\in U$
