Let $f\colon (a,b)\to \mathbb{R}$ be nondecreasing and continuous. If $E=\{x\in (a,b)\mid f'(x)\text{ exists and } f'(x)=0\}$, then $\lambda(f(E))=0$

I need help to understand the proof below of the following theorem.

Let $f\colon (a,b)\to \mathbb{R}$ be an arbitrary function. If $E=\{x\in (a,b)\mid f'(x)\text{ exists and }f'(x)=0\}$, then $$\lambda(f(E))=0.$$$\lambda$ denotes the Lebesgue measure.

You pointed out this wonderful answer to me:

More precisely, let $f:\mathbb R\to\mathbb R$ be an arbitrary function, $\Sigma$ is the set of $x\in\mathbb R$ such that $f'(x)$ exists and equals 0. Then $f(\Sigma)$ has measure 0.By countable subadditivity of measure, we may assume that the domain of $f$ is $[0,1]$ rather than $\mathbb R$. Fix an $\varepsilon>0$. For every $x\in\Sigma$ there exists a subinterval $I_x\ni x$ of $[0,1]$ such that $f(5I_x)$ is contained in an interval $J_x$ with $m(J_x)<\varepsilon m(I_x)$. Here $m$ denotes the Lebesgue measure and $5I_x$ the interval 5 times longer than $I_x$ with the same midpoint. Now by Vitali's Covering Lemma there exists a countable collection $\{x_i\}$ such that the intervals $I_{x_i}$ are disjoint and the intervals $5I_{x_i}$ cover $\Sigma$. Since $I_{x_i}$ are disjoint, we have $\sum m(I_{x_i})\le 1$. Therefore $f(\Sigma)$ is covered by intervals $J_{x_i}$ whose total measure is no greater than $\varepsilon$. Since $\varepsilon$ is arbitrary, it follows that $f(\Sigma)$ has measure $0$.

This answer is certainly correct, but for me, who am just a student, it is too lacking in details that I cannot write. I can't translate what is said into symbols. Could someone be kind enough to explain the details of this answer to me?

Prior state of this question.

I was looking for some text or some suggestions to prove this statement using Vitali's lemma, but at first, under the additionalnondecreasingness and continuity assumptions for $f$, hoping the proof should prove to be more simplified.I even decided to open a bounty, in the hope that someone will be able to provide me with a proof of this fact with these hypotheses using Vitali's Lemma. But since I didn't receive any answer, I shifted my attention to the more general result.