The proof of no measurability of Vitali set assumes that the measure defined on $2^{\mathbb{R}}$ is translate invariant. This is not a condition that a general measure has to satisfy. Furthermore, the proof didn't use more facts than the ones mentioned in the sigma algebra for the sets in which the measure is defined to get a contradiction. So I don't understand why is it not possible to define a measure on $2^{\mathbb{R}}$?
What I understand from the introduction to measure theory is that we want to define the notion of a measure for subsets of a universal set. But we cannot do that if we want to impose some properties on the measure function. And Vitalli set is an example. So we need to formalize the properties for a collection of subsets in which a proper measure could be defined on the subsets of the universal set. And we arrived at the definition of a sigma algebra? But the power set of any set is a sigma algebra no?
I don't understand how is the definition of sigma algebra help with defining a measure?
You are correct. The construction of the Vitali set only shows that there is no countably additive translation invariant measure on the power set of the reals such that the unit interval has finite nonzero measure. In particular, it shows that the Lebesgue measure (which is translation invariant) cannot measure all subsets of the reals.
It is well-known that there is a finitely additive translation invariant extension of the Lebesgue measure to all subsets of reals. Finally, the zero measure and the counting measure are translation invariant and countably additive, but the intervals don't have finite nonzero measure with respect to those.
You cannot prove that there is an extension of the Lebesgue measure to all subsets of the reals (or, equivalently, that there is a countably additive measure such that the intervals have finite nonzero measure), and you are unlikely to be able to prove that there is no such extension. The reason for this is that the existence of such measures is equivalent to the existence of certain large cardinals smaller than the continuum, and if ZFC is consistent, it is also consistent with the statement that such cardinals do not exist. It is also believed to be consistent with the statement that such cardinals do exist.
About the rest of your question: I don't really understand what you are asking about. If you still have some other question, please ask a separate question (do try to make it focused).