To expand on my comment:
One approach is just to invoke the Weierstrass approximation theorem. This works even if $f$ is merely continuous, and it gives a $g$ which is a polynomial, which is drastically stronger than just being Lipschitz or even $C^\infty$.
You could also give a more direct proof. An absolutely continuous function has a derivative which is $L^1$; a Lipschitz function has a derivative which is bounded. You can approximate $L^1$ functions by bounded functions. Now, to get from the derivative back to the function, what could you do...?
Indeed, integrate. So find a bounded measurable function $h$ which is close to $f'$ in $L^1$ norm. What can you say about the difference between the integrals (from $a$ to $x$) of $f'$ and $h$?
So we can get $\int_a^x f'(t) dt - \int_a^x h(t)dt$ to be small, right? Or in other words, we can get $f(x) - f(a) - \int_a^x h(t)dt$ to be small. So what if we set $g(x) = f(a) + \int_a^x h(t)dt$?