Question about a proof of Second Derivatve Test 
Partial Statement: Let $f$ be defined in a neighborhood of $c$ with $f'(c) = 0$. Then $f''(c) > 0 \implies f(c)$ is a local minimum of $f$


Proof: Suppose  $f$ is defined in a neighborhood of $c$ with $f'(c) = 0, \ f''(c) > 0 $. $\color{red}{\text{It can be shown}}$ there's punctured neighborhood of $N = (c - r, c) \cup (c, c + s)$ of $c$ s.t. $\displaystyle{\frac{f'(x) - f'(c)}{x - c} > 0}$ for $x \in N$ meaning $f'(x) < 0$ for $x \in (c - r, c)$ and $f'(x) > 0$ for $x \in (c, c + s)$. It follows $c$ is a local minimum as $f$ dips towards $c$ from the left and grows back up on the right of $c$.

My question is about the $\color{red}{\text{part in red of the quote above}}$. How does that follow? Here below is my attempt. If incorrect, what would be a better way to show that? Thanks.
By definition, $f$ is differentiable at $c$ if $\displaystyle{\forall\varepsilon > 0, \exists \delta>0 \ \left|\frac{f(x) - f(c)}{x - c} - f'(c)\right| < \varepsilon}$ for any $x$ with $0<|x - c| < \delta.$ Then $\displaystyle{\frac{f(x) - f(c)}{x - c} = f'(c)} > 0$. Applying the definition above one more time we get $\displaystyle{\frac{f'(x) - f'(c)}{x - c} = f''(c)} > 0$.
 A: Per the part in red:
We have $f''(c) > 0$, or
$$\lim_{x\to c} \frac{f'(x) - f'(c)}{x-c} = f''(c) > 0$$
Let's use $\frac{f''(c)}{2}$ as our $\varepsilon$. There is some $\delta$ such that for all $x$ if
$$0 < |x - c| < \delta \implies \left|\frac{f'(x) - f'(c)}{x-c} - f''(c)\right| < \frac{f''(c)}{2}$$
or
$$-\frac{f''(c)}{2} < \frac{f'(x) - f'(c)}{x-c} - f''(c)< \frac{f''(c)}{2} $$
$$\frac{f'(x) - f'(c)}{x-c} > \frac{f''(c)}{2} > 0$$
In order for the LHS to be $> 0$, for $x>c$ we must have $f'(x) > f'(c) = 0$.
Similarly, for $x < c$ we have $f'(x) < 0$.
Thus, $f$ is decreasing on some interval to the left of $c$ and increasing to the right, and so, $c$ is a local minimum.
A: A prerequisite is missing: $ f$ should be a $C^2-$ function.
From $f''(c)>0$ and the continuity of $f''$ we get a punctured neighborhood $N = (c - r, c) \cup (c, c + s)$ such that $f''(t) >0$ for all $t \in N.$
If $x \in N$, then there is, by the mean value theorem $t \in $ such that
$$\frac{f'(x) - f'(c)}{x - c}=f''(t).$$
Hence
$$\frac{f'(x) - f'(c)}{x - c} >0.$$
