To prove that Cantor function is not absolute continuous, do we need to take countable infinite number of subintervals? This highly upvoted answer claims that we can show that the Cantor function is not absolute continuous by taking a finite number of subintervals. However, one comment right below the answer claims that the "answer" is incorrect (insufficient) because based on Royden, we need to take countable infinite number of subintervals. The original text of Royden is attached here. I am not sure if the comment is correct. What is the complete proof?

 A: The commenter in the link you gave was confused. Absolute continuity only trades in finite collections of intervals, not infinite ones. In fact, the answer in the link is exactly the same as in royden.
A: The comment is incorrect.  The Cantor set can indeed be covered by finite unions of open intervals with arbitrarily small total length, as Elchanan Solomon explained in his response to the comment.  
In detail, the Cantor set $C$ is defined as the intersection of a sequence of sets $C_n$, where each $C_n$ is a finite union of closed intervals of total length $(2/3)^n$.  Taking slightly larger open intervals around each of these closed intervals, we can find a finite union of open intervals $U$ which contains $C_n$ and has total length at most $(2/3)^n+\epsilon$ for any given $\epsilon>0$.  Taking $n$ to be large enough and $\epsilon$ to be small enough, $(2/3)^n+\epsilon$ will be smaller than any given $\delta>0$.  Thus for any $\delta>0$, there exists a finite union of open intervals $U$ containing $C$ with total length less than $\delta$.
Alternatively, following the argument in Royden that you attached yourself, you can actually just use the interior of $C_n$ as your open intervals.  These open intervals don't quite cover $C$, but they are still good enough to prove the Cantor function is not absolutely continuous, since the Cantor function is still constant on each component of their complement.
