What I meant was that that we can pick any sequence of x's such that q(x) != q(x0) for all the x's in that sequence.

Maybe the word any was a bit confusing. I wanted to say that we compute the derivative for any ONE sequence, and then the assumption of differentiability of p(q(x)) says the derivative must be the same when calculated on ANY other sequence (as the derivative's limit exists).

I'm not sure we need Lipschitz condition here though.

I do admit that the assumption for p(q(x)) to be differentiable somewhat limits our proof, as chain rule should also prove that p(q(x)) is differentiable, besides the formula itself. Though the complete proof requires a bit more technical results, I don't plan to add it.