Proof that a continuous function with continuous right derivatives is differentiable.

I am looking at the proof given on this question given by "23rd". Continuous right derivative implies differentiability

I asked in a comment a question about 5 months ago, but didn’t recieve a reply, so I am asking here instead. I have one statement in the proof I dont understand, I have marked it bold, could you please explain it to me? It is the statement "Then from $cin E_delta $ and $mle g(c)le M$ it is easy to see that $c<b$ is impossible".

Here is what the person who asked the question wanted to prove:

Let $f: [0, infty) rightarrow mathbb{R}$ be continuous and have right derivatives at each point in the domain, with the right derivative function being continuous. Then $f$ is differentiable.
Let us denote the right derivative of $f$ by $g$.

Here is the proof by "23rd":

Let us denote the right derivative of $f$ by $g$.

Lemma: Given $a<b$ and $mle M$, if $mle gle M$ on $[a,b]$, then $$mlefrac{f(b)-f(a)}{b-a}le M.$$

Proof: Define
$$L(a)=g(a)quadtext{and}quad L(x)=frac{f(x)-f(a)}{x-a}, xin(a,b].$$
By definition, $L$ is continuous on $[a,b]$, and it suffices to show that for every $delta>0$,
$$E_delta:=Big{xin[a,b],Big|, m-deltale L(y)le M+delta, forall yin[a,x] Big}=[a,b].$$
By definition and the continuity of $L$, we know that $E_delta=[a,c]$ for some $cin[a,b]$, and from $mle g(a)le M$ we know $c>a$. Then from $cin E_delta $ and $mle g(c)le M$ it is easy to see that $c<b$ is impossible. Therefore, $c=b$ and the lemma follows. $quadsquare$

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Now let us show that $f$ is differentiable for any $x>0$. Since $g$ is continuous, given $0<h<x$, we can define
$$m_h=min_{yin[x-h,x]}g(y),quad M_h=max_{yin[x-h,x]}g(y),$$
and we know that
$$lim_{hto 0^+}m_h=lim_{hto 0^+}M_h=g(y).$$
Due to the lemma, for $a=x-h$, $b=x$, $m=m_h$ and $M=M_h$, we have
$$m_hlefrac{f(x-h)-f(x)}{-h}le M_h.$$
Let $hto 0^+$, it follows that the left derivative of $f$ at $x$ exists and is equal to $g(x)$, i.e. $f$ is differentiable at $x$. $quadsquare$