I believe Lebesgue's proof was nonconstructive, and went as follows. There are at most continuum many countable ordinals, and only continuum many functions in each Baire class (exercise - how many countable subsets does a set of size continuum have?), so only continuum many Baire functions total. But there are two to the continuum many functions. (Note the similarity with the proof that there is a non-Borel set of reals, even a measurable one!)
Meanwhile, I don't think a constructive proof can truly exist, since it is consistent with ZF that every function is Baire. This is the case in models where the reals are a countable union of countable sets.
That said, here is a sorta-example: biject the Baire functions with the reals and diagonalize. That is, let $f $ be a subjection from the reals to the Baire functions, and let $g (r)=f (r)(r)+1. $