Consequence Vitali covering lemma

Let $f\colon[a,b]\to\mathbb{R}$ be a real valued function and put$$A=\{x\in(a,b)\colon f\text{ is differentiable at }x,\; f'(x)=0\}.$$

Let $\lambda$ denote the Lebesgue measure on $[a,b]$, then the following holds:

1. Let $\varepsilon>0$, $\alpha>0$ and $y\in f[A]$, then there exists an $x\in A$ and $\delta>0$ such that$$\forall a\in(x-\delta,x+\delta)\colon f(a)\in[y-\delta\varepsilon, y+\delta\varepsilon]$$ and $$\lambda\big([y-\delta\varepsilon, y+\delta\varepsilon]\big)<\alpha.$$
2. Vitali's covering lemma implies that $\lambda^*(f[A])=0$, where $\lambda^*$ denote the Lebesgue outer measure.

Vitali's Covering lemma is stated as

Let $E$ be a set of finite Lebesgue outer measure and $\mathcal{F}$ a collection of intervals the that $E$ in the sense of Vitali. Then, for all $\varepsilon>0$ there exists a finite disjoint collection $\{I_1,\ldots, I_n\}$ such that $$\lambda^*\left(E\setminus\bigcup_{k=1}^nI_k\right)<\varepsilon.$$

I was wondering if anyone knows how to prove these two statements. Thank you in advance.