You can think of absolute continuity as motivated by / modeled on "differentiable almost everywhere" plus "satisfies Fundamental Theorem of Calculus".
Precisely, $f$ is absolutely continuous if and only if $f$ is differentiable almost everwhere and $f(x) = f(a) + \int_a^x f'(x) dx$ for all $x \in [a,b]$.
At first glance, it may seem like a.e.-differentiability should be a nice enough property to ensure FTC is true, but there are counterexamples (like the Cantor function). You can think of absolute continuity as a way of shoring up that kind of pathology, i.e. it eliminates so-called singular (in the measure-theory sense) functions.
The "disjoint" part of the definition serves to weaken the definition a little bit. If "disjoint" were omitted, you'd be describing a condition for Lipschitz functions, which is stronger than needed (if your aim is "abs. cont." $\iff$ "a.e.-diff + FTC"). That is, there are functions which are abs. continuous, but not Lipschitz (like $\sqrt{x}$ on $[0,1]$).
As an exercise, show that this definition fails for $\sqrt{x}$ on $[0,1]$ if the disjoint hypothesis is removed. Hint: take each of the intervals $[a_k,b_k]$ to be the $\textit{same}$ interval $[0,\alpha]$ for some appropriately small $\alpha$.
Here is a proof that uses properties of the integral.
Note that $\sqrt x= \int_0^x \frac{1}{2\sqrt{t}} dt$ for all $x\in [0,1]$. Also
$$\sqrt{x}-\sqrt{y}= \int_y^x \frac{1}{2\sqrt{t}} dt$$
Thus, if $(a_i,b_i) \subset [0,1],i=1,\dots,N$ are disjoint intervals, then
$$\sum_{i=1}^N |\sqrt{b_i}- \sqrt {a_i}|= \int_{\bigcup_{i=1}^N (a_i,b_i)} \frac{1}{2\sqrt{t}} dt $$
Since the function $g(t)= 1/2\sqrt{t}$ is integrable, for each $\varepsilon>0$ there exists $\delta>0$ such that for all sets $A$ with measure smaller than $\delta$ we have
$$\int_A g(t)dt <\varepsilon$$
Note that $A:= \bigcup_{i=1}^N (a_i,b_i)$ has measure less than $\delta$ if and only if $\sum_{i=1}^N |b_i-a_i|<\delta$ because the intervals are disjoint.
We note that the integrability of the derivative $g$ is important here. It is not true that if $f$ is absolutely continuous on every interval $[\varepsilon,1]$ and continuous at $0$, then it is absolutely continuous on $[0,1]$. One example of such a function is $f(x)=x\sin(1/x)$. This is absolutely continuous on every interval $[\varepsilon,1]$ because it has bounded derivative there, thus it is Lipschitz. However, it is not absolutely continuous on $[0,1]$, since it is not even BV there.
Best Answer
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 $\delta = \epsilon^2.$