necessity of semi finiteness As stated in this question sigma finite measures are semi finite. 
$\sigma$-finite measure and semi-finite measure
I am interested in the question if we can weaken the sigma finiteness condition to uncountable sets, i.e. suppose $(\Omega, F,\mu) $ is a measure space and $G\subset F$ exists s.t. $\mu (A)<\infty$ for all $A\in G$ and $\bigcup_{A\in G}A=\Omega $ I believe it is true, but was not able to proof it.

It's not true as stated:
Let $\Omega$ be uncountable with $\mu(\Omega) =\infty$ and $\mu(\{\omega\})=0$ for each $\omega\in\Omega$. Let $F$ be the smallest $\sigma$-algebra generated by the singletons, i.e. it consists of countable subsets and their complements. 
Finally, choose $G:=\{\{\omega\} \, :\, \omega\in\Omega\}$.
I guess, a similar but more sophisticated counterexample can be cooked up for the case when $\mu(A)>0$ is also assumed for some (or all) $A\in G$. 
