If $f$ is Lipschitz, we know by Rademacher's theorem that it is differentiable a.e. That is, the set of non-differentiable points is of measure zero. But this does not necessarily imply that it is a countable set.
Can we find a Lipschitz continuous function that has an uncountable differentiable points?
Yes, from the paper below
we know that for a given set $E$, there exists a Lipschitz function such that $f$ is not differentiable at $E$ if and only if $E$ is of Lebesgue null. It is easy to find a uncountable set of Lebesgue null, say Cantor set.