Does the everywhere differentiability of $f$ imply it is absolutely continuous on a compact interval?

No. Consider $f(x) = x^2 \sin(1/x^4)$ on $[-1,1]$. Note that for $\beta_n = (\pi n)^{-1/4}$ and $\alpha_n = (\pi (n+1/2))^{-1/4}$ we have $\beta_n - \alpha_n = \Theta( n^{-5/4})$ while $|f(\beta_n) - f(\alpha_n)| = \Theta(n^{-1/2})$.