The question basically says it all. It is a well-known result that there exists a model $ \mathcal{M} $ of ZF with the property that $ \mathbb{R}^{\mathcal{M}} $ (here, $ \mathbb{R}^{\mathcal{M}} $ is the element of $ \mathcal{M} $ that $ \mathcal{M} $ thinks is its real-number field) is the countable union of countable sets. How can this be compatible with the assertion that the unit interval $ [0,1]^{\mathcal{M}} $ (the object that $ \mathcal{M} $ thinks is the unit interval) has measure $ 1 $? Herein lies the problem. Every countable set has measure $ 0 $, hence a countable union of countable sets is a countable union of measure-$ 0 $ sets, which, in turn, implies that $ \mathbb{R} $ has measure $ 0 $. It then follows that $ [0,1] $ has measure $ 0 $.
Unless some form of the Axiom of Choice (AC) is needed to make sense of measure theory (i.e., some variant of AC is required to ensure that the standard Borel measure $ \mu $ exists in the first place), I do not see any way to resolve this issue.