Question about Vitali Covering (from a Lemma in Royden and Fitzpatrick's book) 
Definition. For a real valued function $f$ and an interior point $x$ of its domain, the upper derivative of $f$ at $x$ denoted by $\overline{D}f(x)$ is defined as follows: $$\overline{D}f(x)=\lim_{h\rightarrow0}\left[ \sup \left \{\frac{f(x+t)-f(x)}{t}: 0<|t|\leq h \right \} \right]$$

A Lemma in Royden and Fitzpatrick's Real Analysis book says:
Lemma. Let $f$ be an increasing function on the closed, bounded interval $[a,b]$. Then for each $\alpha>0$, $$m^*\{x\in (a,b) : \overline{D}f(x) \geq 
\alpha \} \leq \frac{1}{\alpha}[f(b)-f(a)].$$
The book proceeds to prove this by:

Let $\alpha>0$. Define $E_{\alpha}:=\{x\in (a,b): \overline{D}f(x)\geq\alpha \}$. Choose $\alpha' \in (0,\alpha)$. Let $\mathscr{F}$ be the collection of closed, bounded intervals $[c,d]$ contained in $(a,b)$ for which $f(d)-f(c)\geq \alpha ' (d-c)$. Since $\overline{D}f\geq \alpha$ on $E_{\alpha}$, $\mathscr{F}$ is a Vitali covering for $E_{\alpha}$.

I can follow the rest of the proof, but why is the statement above true? In particular, why is $\mathscr{F}\neq \emptyset$, and why is it a Vitali covering for $E_{\alpha}$? I have an inkling that it might be due to the fact that $f$ is increasing in $(a,b)$ and thus it can only have a countable number of discontinuity on it, but I can't quite get a solid grasp of it.

Take any $x \in E_\alpha$. Now, since $\overline{D}f(x)\geq\alpha$, it follows that for some small $\delta$, $t<\delta \implies\frac{f(x+t)-f(x)}{t}\geq\alpha'$.
(The definition for the upper derivative above is slightly wrong, I will edit it)
Putting $t=d-x$, this means that $t<\delta \implies f(d)-f(x) \geq \alpha'(d-x)$. The interval $[d,x]$ is in $\mathscr{F}$ for every $d$ close enough to $x$,for arbitrary $x$ in $E_\alpha$. This makes $\mathscr{F}$ a Vitali covering for $E_\alpha$.
