A different approach might help your thinking on this. Locally what does a derivative do? If the value $f'(x_0)=\pm 5$, then closed intervals $J$ close to
that point get expanded in length by a factor of $5$, i.e,. $f(J)$ is about five times the length of $J$. At least that is roughly how we would explain "what derivatives do" to an elementary student. (We would have to sell the idea that, for continuous functions, $f$ maps a closed interval $J$ to a closed interval $f(J)$.)
At the level of a course in measure theory we shouldn't forget our baby intuition. It is not difficult with the usual apparatus of the Lebesgue outer measure to prove this lemma, which is just the same kind of "expansion" idea, except that rather than "intervals" expanded in length "sets" get expanded in measure.
Lemma. Let $F:\mathbb R\to \mathbb R$ be an arbitrary function that has a derivative at every point of a set $E$ and suppose that
$|f'(x)|\leq M$ for every $x\in E$. Then
$$\lambda^*(f(E))\leq M \lambda^*(E).$$
[Aside: Actually you don't need the existence of a derivative. It is enough if $-M \leq D_- f(x) $ and $D^+ f(x)\leq M$ where these are, respectively the lower left Dini derivative and the upper right Dini derivative. But assuming a derivative makes it marginally easier to construct a proof.]
Now your problem is a trivial consequence of the lemma, without any fussing over continuity of the derivative and the mean-value theorem which are quite irrelevant to the nature of the observation. If $f'(x)=0$ for
all $x\in E$ then $|f'(x)|<\epsilon$ for any $\epsilon>0$ and so
$$\lambda^*(f(E))\leq \epsilon \lambda^*(E).$$
It follows that, if $E$ is bounded, then $\lambda^*(f(E))=0$. For the unbounded case just use a sequence $E\cap [-n,n]$.
Note that measurability of $f(E)$ then follows immediately since it has Lebesgue outer measure zero.
