Proof that $\sqrt x$ is absolutely continuous.

Without measure theory:

Lemma: Suppose $0<a<b$ and $0\le h\le a.$ Then

$$\sqrt {b} - \sqrt {a}\le \sqrt {b-h}-\sqrt {a-h}.$$

On the right, both $a,b$ have been shifted to the left by $h.$ A glance at the graph of $y=x^{1/2}$ makes the result intuitively clear: The rate of growth of $\sqrt {x}$ is larger as $x$ gets smaller. To prove the lemma, define $f(h)$ to be the right side minus the left side. Then $f$ is continuous on $[0,a],$ differentiable on $(0,a],$ and $f(0)=0.$ Verify $f' > 0$ on $(0,a]$ to then give the result.

Suppose now that $[a_1,b_1] < [a_2,b_2] < \cdots < [a_n,b_n]$ are pairwise disjoint. Claim:

$$\tag 1 \sum_{k=1}^{n} (\sqrt {b_k}-\sqrt {a_k}) < \sqrt {\sum_{k=1}^{n} (b_k - a_k)}.$$

Proof: By the lemma, the sum on the left can only increase if we shift all of the $[a_k,b_k]$ to the left so they're right next to each other. We then obtain intervals $[c_{k-1},c_k],$ with $0=c_0 < c_1 < \cdots <c_n,$ and lengths $c_k-c_{k-1} =b_k-a_k.$ Thus

$$ \sum_{k=1}^{n} (\sqrt {b_k} - \sqrt {a_k}) \le \sum_{k=1}^{n} (\sqrt {c_k} - \sqrt {c_{k-1}}) = \sqrt {c_n}.$$

But $c_n$ is just the sum of the lengths of these intervals, so we have $(1).$

Now it's easy to find a $\delta$ for a given $\epsilon$ in proving absolute continuity of $\sqrt x:$ Just choose $\delta = \epsilon^2.$

Note that $\sqrt{x}$ is differentiable with derivative $\frac{1}{2\sqrt{x}}$. Since: $$\sqrt{x} = \int_0^x \frac{{\rm d}t}{2\sqrt{t}}$$and the derivative is Lebesgue-integrable (do note that the above integral is improper, though), $\sqrt{x}$ is absolutely continuous.