"Partitioning" an uncountable set "equally" I just observed a simple fact about the real numbers, that if U is an uncountable subset of the set of real numbers, then there exists a real number r, such that both the set V  of all elements of U less than r and the set W of all elements of U greater than r, are uncountable. In the proof, I however needed the completeness axiom of the set of real numbers, but not its field structure. My first question is that, is the same true for any totally ordered set (without the completeness axiom)? My second question is somewhat different. The continuum hypothesis surely guarantees that any two uncountable subsets of the real numbers are bijective. But, without the continuum hypothesis, can we say that there exists a real number r, such that the sets V and W as defined above, are bijective?   
 A: Try the smallest uncountable ordinal $\omega_1$, which is of course a totally ordered set.
We can follow the construction from your question, i.e. for any $r\in \omega_1$ we can define the two sets $V=\{\,x\in\omega_1\mid x<r\,\}$ and $W=\{\,x\in\omega_1\mid x>r\,\}$. Now since $\omega_1$ is an ordinal, $r$ is also an ordinal and is strictly smaller than $\omega_1$. Moreover, $V$ is simply $r$, hence countable. In other words: For the totally ordered uncountable set $\omega_1$, it is not possible to cut it into two uncountable subsets, thus giving a negative answer to your first question.
As Asaf kindly remarks, the order is complete. So we did not only find acounterexample, an uncountable ordered set that does not allow the kind of partitioning you describe - but we notice that completeness is not even the (only) key point!
A: If you can prove that the set of $r$ that are too small is bounded upwards, then the completeness of the real numbers give you an $r$ that works. You don't need the continuum hypothesis for this, just the ability to compare the size of any two subsets of $\Bbb R$.
