The answer is no and it follows from the following: 

>It is consistent that $AC$ fails but for all infinite cardinals $\kappa, 2 \cdot \kappa=\kappa.$

The above result is proved by Sageev:

 Sageev, Gershon [An independence result concerning the axiom of choice](http://www.sciencedirect.com/science/article/pii/0003484375900029). Ann. Math. Logic 8 (1975), 1–184.

In a model as above, every infinite set is splittable but $AC$ fails in it.