We can prove directly.
Choose $M>0$ such that $|f(x)|\leq M$ and $|g(x)|\leq M$ for all
$x\in[a,b]$. Let $\varepsilon>0$ be given. Since $f$ and $g$ are
absolutely continuous, there exists $\delta>0$ such that for any
(fintely many) pairwisely disjoint subintervals $(x_{1},y_{1}),\ldots,(x_{n},y_{n})$
of $[a,b]$, if $\sum_{i=1}^{n}(y_{i}-x_{i})<\delta$, then $\sum_{i=1}^{n}|f(y_{i})-f(x_{i})|<\frac{\varepsilon}{2M}$
and $\sum_{i=1}^{n}|g(y_{i})-g(x_{i})|<\frac{\varepsilon}{2M}$. Now,
let $(x_{1},y_{1}),\ldots,(x_{n},y_{n})$ be pairwisely disjoint subintervals
of $[a,b]$ with $\sum_{i=1}^{n}(y_{i}-x_{i})<\delta$. For each $i$,
observe that
\begin{eqnarray*}
& & \left|f(y_{i})g(y_{i})-f(x_{i})g(x_{i})\right|\\
& \leq & \left|f(y_{i})g(y_{i})-f(x_{i})g(y{}_{i})\right|+\left|f(x_{i})g(y_{i})-f(x_{i})g(x_{i})\right|\\
& \leq & M\left|f(y_{i})-f(x_{i})\right|+M\left|g(y_{i})-g(x_{i})\right|.
\end{eqnarray*}
It follows that
\begin{eqnarray*}
& & \sum_{i=1}^{n}\left|f(y_{i})g(y_{i})-f(x_{i})g(x_{i})\right|\\
& \leq & M\sum_{i=1}^{n}\left|f(y_{i})-f(x_{i})\right|+M\sum_{i=1}^{n}\left|g(y_{i})-g(x_{i})\right|\\
& < & \varepsilon.
\end{eqnarray*}
Therefore $fg$ is absolutely continuous.