A quick note that (2) doesn't follow from ZF (assuming ZF is consistent) or even ZF+DC.  

(2)+Countable Choice for Finite Sets (CC(fin)) implies that every uncountable set has a non-principal (finitely additive, non-trivial) measure.  Let $\Omega$ be uncountable.  Let $\mu$ be as in (2).  Let $A = \{ x : \mu(\{ x \}) > 0 \}$.  The standard proof that a finite measure has only countable many atoms only uses CC(fin) (there are at most $n$ elements $x$ such that $\mu(\{ x \}) \in [1/n, 1/(n-1))$).  So $A$ is countable.  So $\mu(A)<1$.  Then $\nu(B) = \mu(B-A)$ defines a non-principal measure and is non-trivial since $\nu(\Omega)=1-\mu(A)>0$.

But Blass gave a model of ZF+DC with no non-principal measures ("A model without ultrafilters", Bull. Acad. Polon. Sci. 25 (1977), 329–331; I haven't read the paper, but Pincus and Solovay say that it proves this).  Since DC implies CC(fin), (2) is false in that model.

A different way to see that (2) doesn't follow from ZF is this.  Take a model where there is an infinite Dedekind-finite set but CC(fin) still holds (the [Consequences of AC website][1] lists a bunch of models where Form 10 is true and Form 9 is false).

Now if $\Omega$ is an infinite Dedekind-finite set and $\mu$ satisfies (2), then every point of $\Omega$ will have to be an atom of $\mu$ (or else $\mu(\Omega - \{ x \}) = \mu(\Omega)$).  But by CC(fin), the set of atoms is countable, which is impossible for a Dedekind-finite set.

I'm curious if (2) implies CC(fin).

  [1]: http://consequences.emich.edu/conseq.htm