Q:

Partition the set of positive real number into pairs without axiom of choice?

Backround. Yesterday I made a comment on Achuille hui's answer and then I discuss on the chat with Paul Plummer about the way to partition the set of positive real number into pairs $\Bbb{R}^+=\bigcup_{i\in I}\{u_i,v_i\}$. It seems that the axiom of choice is 'needed'. Unfortunately I don't feel comfortable with it. 

Question: Is it possible to avoid axiom of choice to construct such a partition ? 

A:

No axiom of choice is needed.
First of all, being able to partition something into pairs is not dependent on the actual set, just its cardinality. Namely, if we can partition $\mathcal P(\Bbb N)$ into pairs, and we can prove there is a bijection between $\Bbb R^+$ and $\mathcal P(\Bbb N)$, then it gives us a partition of $\Bbb R^+$ into pairs.
And indeed we can do both these things without the axiom of choice. Partitioning $\mathcal P(\Bbb N)$ is easy. Simply consider $\{A,B\}$ such that $A\cup B=\Bbb N$ and $A\cap B=\varnothing$. Showing there is a bijection between the two sets is also not difficult, by noting that the proof of the Cantor-Bernstein is not using the axiom of choice, and that Dedekind cuts and the Cantor set define the two needed injections.
(Alternatively, you can just partition $\Bbb R\setminus\{0\}$ into pairs in the obvious way, and use the fact that $\Bbb R^+$ and $\Bbb R\setminus\{0\}$ have the same cardinality again, which might be a bit easier to prove.)

Having said that, it is consistent that there are sets which cannot be partitioned into pairs. For example in Cohen's first model there exists an infinite Dedekind-finite set of reals.
Little known is the fact that in that model, given that set, every partition into finite parts is co-finitely made of singletons. Therefore you cannot partition this set into pairs.

