I think this is a counter-example.
Let $C$ be a cantor set of positive measure, so $C$ is nowhere dense, perfect and is the countable decreasing intersection of sets $C_{n}$ each of which are a finite union of closed disjoint intervals in $[0,1]$.
Let $f(x)=\int_{0}^{x}\chi_{C}(t)\mbox{ }dt,$ so $f$ is certainly absolutely continuous.
Assuming I did this right $f(C)$ contains $[0,m(c)).$ Consider each $C_{n},$ since $f$ is increasing and continuous $f([0,1])$ contains $[0,m(C)].$ Also $[0,1]\setminus C_{n}$ is a union of open intervals each of which are in the compliment of $C$ so it follows that $f$ is constant on each such interval. Since $f([0,1])=[0,m(C)]$ it follows that for any $x\in [0,m(C))$ we can find $t\in C_{n}$ so that $f(t)=x.$ Indeed we already know we can do this with $t\in [0,1]$ but since $f$ is constant on the intervals in the complement of $C_{n}$ we can force $t\in C_{n}$ (for instance if $x$ is in some interval $I$ in $[0,1]\setminus C_{n}$ then its left endpoint is in $C_{n}$ and since $f$ is constant on $I$ we have that $f$ has the same value at the left-endpoint of $I$ as on $I$.)
Now fix $y\in [0,m(C)).$ Since $\lbrace x\in C_{n}:f(x)=y\rbrace$ is non-empty and these sets are decreasing, (since the $C_{n}$ are decreasing) by compactness we can find $x\in C$ so that $f(x)=y.$ Thus $f(C)=[0,m(C))$ and we have found a nowhere dense set which is mapped to a set which is not nowhere dense.
This is impossible. Baire proved that if a function defined on $\mathbb R$ is of Baire class 1, then it is continuous everywhere except, possibly, for a meagre set. And by another Baire's theorem a complement of a meagre set in $\mathbb R$ is dense.
An elementary exposition of this and related results can be found in the nice little book by Oxtoby.
Edit. Another good reference which covers Baire's theorems and provides some historical background is "The calculus gallery" by W. Dunham.
Best Answer
It is a theorem due to Blumberg (New Properties of All Real Functions - Trans. AMS (1922)) and a topological space $X$ such that every real valued function admits a dense set on which it is continuous is sometimes called a Blumberg space.
Moreover, in Bredford & Goffman, Metric Spaces in which Blumberg's Theorem Holds, Proc. AMS (1960) you can find the proof that a metric space is Blumberg iff it's a Baire space.