In Foundations of Modern Probability, Kallenberg, (2nd edition) page 7, lemma 1.13 the following proof is given:
I can understand how the proof goes through if we assume that the measure space $(S,\pmb{S})$ is a standard Borel space (that is, a polish space with the Borel sigma algebra corresponding to the polish topology) , since then we may apply Kuratowski's theorem which says that $(S,\pmb{S})$ is Borel isomorphic to $(F = \{x_1,...,x_k\}, 2^F)$ or $(\mathbb{N},2^{\mathbb{N}})$ in which case we can prove the theorem directly for these cases, and if it is uncountable it is borel isomorphic to $([0,1],\mathcal{B}([0,1]))$ and we may proceed as Kallenberg does, but if $(S,\pmb{S})$ is merely Borel isomorphic to a Borel subset of $[0,1]$ then I do not see that we can ensure it is standard Borel, and so Kuratowksi's theorem may not apply, so how can we further reduce to the case when $S = [0,1]$ as Kallenberg suggests?