Let $f: [a,b] \rightarrow \mathbb{R}$ be a function with the following properties: To every $\epsilon > 0$ there exists a $\delta > 0$, such that every finite system $(\mathopen{]}a_j,b_j\mathclose{[})_{1 \leq j \leq n }$ of not necessarily pairwise disjoint open sets $\mathopen{]}a_j,b_j\mathclose{[} \subseteq [a,b]$ implies $\sum_{j=1}^n (b_j - a_j) \leq \delta \implies \sum_{j=1}^n |f(b_j) - f(a_j)| \leq \epsilon$.
Show that $f$ is Lipschitz continuous.
At the beginning, I had no idea what to do. After research, I found a post here that had a similar problem in it, but there was an additional requirement to $f$. $f$ needed to have a bounded derivative which is not the case here (is it?). So, still no idea what to do.
The answer to the title question is "No, absolute continuity does not imply Lipschitz continuity". This is shown for example by the square root function on $[0,1]$. Of course we can find absolutely continuous functions with much worse behaviour, say that the derivative is unbounded on every nonempty open interval contained in the domain.
But the conditions you are given for $f$ are — very sneakily, I overlooked it at first too — stronger than just absolute continuity. The key point is that the intervals $\mathopen{]} a_j, b_j\mathclose{[}$ are not necessarily disjoint.
This changes the game. Let $\delta_1$ be the $\delta$ corresponding to $\varepsilon = 1$, i.e. we have $$\sum_{j = 1}^n (b_j -a_j) \leqslant \delta_1 \implies \sum_{j = 1}^n \lvert f(b_j) - f(a_j)\rvert \leqslant 1$$ for all families of open intervals $\mathopen{]} a_j, b_j\mathclose{[}$, $1 \leqslant j \leqslant n$, contained in $[a,b]$.
Now consider $x,y \in [a,b]$ with $0 < y-x \leqslant \delta_1$ and let $n = \bigl\lfloor \frac{\delta_1}{y-x}\bigr\rfloor$. Taking the family of open intervals with $a_j = x$ and $b_j = y$ for $1 \leqslant j \leqslant n$ shows $$\sum_{j = 1}^n \lvert f(b_j) - f(a_j)\rvert = n\lvert f(y) - f(x)\rvert \leqslant 1\,,$$ thus $$\lvert f(y) - f(x)\rvert \leqslant \frac{1}{\bigl\lfloor \frac{\delta_1}{y-x}\bigr\rfloor} \leqslant \frac{1}{\frac{1}{2}\frac{\delta_1}{y-x}} = \frac{2}{\delta_1}(y-x)\,.$$ From this it follows that $$\lvert f(y) - f(x)\rvert \leqslant \frac{2}{\delta_1}\lvert y-x\rvert$$ for all $x,y \in [a,b]$, i.e. that $f$ is Lipschitz continuous with Lipschitz constant $2/\delta_1$. (With a bit more work one can show that $1/\delta_1$ works as a Lipschitz constant for $f$.)