Existence of non-measurable sets requires the use of Axiom of choice. And one can construct models eg. Solovay model without using the Axiom of choice where there are no non-measurable sets. 
 So, Can one start with the the hypothesis that non-measurable sets exist and then 'prove' the Axiom of choice? 



Existence of Non-measurable sets
The answer to your question is "no". 


Using the technique of forcing one can easily violate the axiom of choice in a variety of ways (essentially, by focusing on "large enough sets") without affecting the fact that there are non-measurable sets of reals. This includes leaving a bit of choice to make sense of the construction of Lebesgue measure and development of its basic properties (the principle of "dependent choice for relations on reals" seems more than enough for this). 


As far as I know, there are no known "choice or choice-like principles" even at the level of the reals known to be equivalent or to reasonably follow from the assumption that there are non-measurable sets. 


(By the way, this is form 93 in "Consequences of the axiom of choice" by Howard and Rubin. The companion [website](http://consequences.emich.edu/CONSEQ.HTM) does not reveal any known implications.)


