If $g'(c) \neq 0$, show $g(x) \neq g(c)$ for all $x \in V_d(c)$ (S.A. pp 144 q5.3.8) 
Let $g : (a, b) \to R$ be differentiable at a point $c \in (a, b)$. $V_d(c) := \{ x \in R : |x - c| < d \}$.
First Case
$g'(c) > 0$: We cannot use the mean value theorem since we only know that the
function is differentiable at a single point. Instead use the definition of the derivative.
Since $0 < g'(c) = \lim_{x \to c} \dfrac{ g(x) − g(c) }{ x − c }$, there is an $e > 0$ such that $g'(c) > e$  

1. Intuition please?
2. Why $g'(c) > e$  ?
3. Please demystify the proof's modus operandi? These steps underneath feel fey and tricky.  

and a $d > 0$ such that $0 < |x − c| < d$  implies
$ |g'(c) - \dfrac{ g(x) − g(c) }{ x − c } | < e \iff -e < \color{seagreen}{g'(c) - \dfrac{ g(x) − g(c) }{ x − c }  < e}  $
$ \iff \color{red}{0 < } \color{seagreen}{g'(c) - e < \dfrac{ g(x) − g(c) }{ x − c } } $.  Ergo $ \color{red}{0 < } \color{seagreen}{\dfrac{ g(x) − g(c) }{ x − c } } \quad (☺)$.
When $0 < x − c < d$, (☺) In order for the fraction to $> 0$, (☺) invokes $\color{seagreen}{g(x) - g(c) > 0}$. 
When $0 < -(x − c) < d \iff -d > x - c > 0,$ so that both numerator and denominator $< 0$, we need $\color{seagreen}{g(x) - g(c) < 0}$. 
In both cases, there's a $d$-neighborhood of $x = c$ in which $-g(x) \neq -g(c)$ for $x = c.$ 
Second Case
$g'(c) < 0$: Let $h(x) = −g(x).$
Then $h'(c) = −g'(c) > 0.$ Thus there's a neighborhood of c where $-g(x) \neq -g(c)$ unless $x = c$. Multiplying by −1 yields $g(x) \neq g(c)$ on the same neighborhood.



4. I can do 5.3.7 but don't understand this. Please demystify and expatiate? 

 A: As for the intuition: since a function is differentiable at $c$, and $g'(c)\neq 0$, $g$ behaves almost like a nonconstant linear function in some neighbourhood of $c$ - on some side of $c$ it "goes up", on the other it "goes down".
As for the proof: since $g'(c)>0$, there exists some $0<e<g'(c)$. The choice is arbitrary, we only want it to satisfy this inequalities. The steps below use a definition of derivative (and so, definition of limit) to show that the difference quotient
$$
\frac{g(x)-g(c)}{x-c}
$$
for $x$ sufficiently close to $c$ is so close to $g'(c)>0$, that it must be greater than $e$, and so, greater than $0$. That's the crux of the proof. Simple algebraic manipulation yields the sought inequality.
As for the sine example: this exercise is supposed to show that, despite the conclusion from  5.3.8, non-zero derivative at a point does not imply local monotonicity, even though initially such statement may seem plausible. So it is a counterexample to such strengthening of 5.3.8.
A: I will try to present the "modus operandi" of the pretty standard proof which has been provided in your question.
The proof proceeds in a sequence of steps starting with the assumption that $g'(c) > 0$ (the case $g'(c) < 0$ being handled similarly). Now the idea is to figure out in detail what does $g'(c) > 0$ imply. For this we first note that $g'(c)$ is a limit of $f(x) = \dfrac{g(x) - g(c)}{x - c}$ as $x \to c$. And this limit is positive.
Now you need to understand that if a function $f(x)$ tends to a positive limit $x \to c$ then it means that the values of the function $f(x)$ will also be positive for values of $x$ near $c$ (think of the weird scenario if this does not happen: "a function taking negative values but tending to a positive limit". Without even thinking in terms of symbols and inequalities this statement sounds weird/non-sensical).
Once you are convinced of the fact mentioned in last paragraph you know that the ratio $f(x) = \dfrac{g(x) - g(c)}{x - c}$ is positive for all $x$ sufficiently near $c$. Now we ask : when is a ratio positive? When both numerator and denominator have same sign. This means that sign of $g(x) - g(c)$ is same as that of $x - c$. This means that if $x - c$ is positive (i.e. $x > c$) then $g(x) - g(c)$ is also positive (i.e. $g(x) > g(c)$). And if $x - c$ is negative (i.e. $x < c$) then $g(x) - g(c)$ is also negative (i.e. $g(x) < g(c)$).
So we see that if $x$ is near $c$ we have either $g(x) > g(c)$ or $g(x) < g(c)$. This is what we had to prove that $g(x) \neq g(c)$ if $x$ lies in a certain neighborhood of $c$ and $x \neq c$ (this is what is meant by saying that $x$ is near $c$ in last few paragraphs).
