I prefer the proof by Caron and Trayner because it shows that the recurrence argument is a direct consequence of Fubini' theorem for Lebesgue measure.
However, I am still looking for a published reference to a textbook in measure theory or to a paper in a journal (simpler than Herbert Federer' book) because the link to Caron & Trayner note at Univ Windsor could be unstable.
It is because the set of zeros of polynomials are discrete finite and hence countable. You can cover this finite set E={x_{i}:i=1,2,...,n} with countably open sets of length ε/(2^{i}), ∀ε>0, i.e., x_{i} is contained in an open interval E_{i}=(x_{i}-(ε/(2^{i+1})), x_{i} + (ε/(2^{i+1}))) with μ(E_{i}) = ε/(2^{i}) so that E=∪₁ⁿE_{i} and
My earlier answer was for a polynomial of a single variable with n zeros. However looking at your message, the set of zeros of a polynomial in n-variable is at most a set of dimension n-1 whose Lebesgue measure in the n-D Euclidean space is zero. Any n-1 submanifold or subset in the n-D Euclidean space is a null set or of Lebegue measure zero.
To see it differently, no non zero polynomial of n-1 variables will vanish on a n-D section of the n-D Euclidean space, as no non zero polynomial of a single variable will vanish on a non degenerate 1-D segment.
I agree with your argument for multivariable polynomial, but the point is that it is not so easy to give an elementary proof that "the set of zeros of a polynomial in n-variable is at most a set of dimension n-1", and that is why I prefer the direct simple proof of Caron and Trayner using Fubini's theorem. A "simple" proof was indeed the aim of my initial question.