$\sqrt{x}$ is absolutely continuous but not Lipschitz on $[0,1]$

Let $[u_j,v_j]$ be pairwise nonoverlapping subintervals. Fix $r>0$ and split at $r$ the one family interval, if any, that crosses $r$; monotonicity makes this split preserve its endpoint increment. The resulting pieces contained in $[0,r]$ contribute at most $\sqrt r$ in total, because their increments telescope after gaps are filled. On every remaining interval, whose left endpoint satisfies $u_j\ge r$, [given] $\sqrt{v_j}-\sqrt{u_j}=\frac{v_j-u_j}{\sqrt{v_j}+\sqrt{u_j}}\le\frac{v_j-u_j}{2\sqrt r}.$

Given $\varepsilon>0$, choose $r>0$ with $\sqrt r<\varepsilon/2$, then require $\sum_j(v_j-u_j)<\varepsilon\sqrt r$. Steps 1.1 and 1.2 make the total endpoint increment less than $\varepsilon$, proving absolute continuity by [L2].

If a Lipschitz constant $K$ existed, the pair $0$ and $1/n^2$ would give $n^{-1}\le K n^{-2}$, hence $n\le K$ for every positive integer $n$, contradicting the Archimedean property. Thus [L3] fails.