$\frac{x-c}{x-y} |\frac{f(x) - f(c)}{x-c} - f'(c)| + \frac{c-y}{x-y}|\frac{f(y) - f(c)}{y-c} - f'(c)| < (\frac{x-c}{x-y} + \frac{c-y}{x-y}) \epsilon$? After substituting the green expression, how do you deduce the last two inequalities below? I'm guessing Triangle Inequality? But after you substitute the green expression, you have three terms, whilst Triangle Inequality has just two variables.

Since $f(x)−f(y)=f(x)\color{red}{−f(c)+f(c)}−f(y)$, a simple
calculation shows that
$\color{green}{\dfrac{f(x) - f(y)}{x-y} = \dfrac{x-c}{x-y}\cdot\dfrac{f(x) - f(c)}{x-c} +\dfrac{c-y}{x-y}\cdot \dfrac{f(y) - f(c)}{y-c}}$
Since both $\dfrac{x-c}{x-y}$ and $\dfrac{c-y}{x-y}$ are positive and sum to 1, it follows that
$\begin{align} 
& \left|\color{green}{\dfrac {f(x)-f(y)}{x-y}}-f'(c)\right| \\
& \le (\frac{x-c}{x-y})\left|\frac{f(x) - f(c)}{x-c} - f'(c) \right| +(\frac{c-y}{x-y})\left|\frac{f(y) - f(c)}{y-c} - f'(c)\right|  \\
& < (\frac{x-c}{x-y} +  \frac{c-y}{x-y})\epsilon \\
& \qquad = \epsilon \end{align}$

 A: Supposing $x<c<y$, we justify it by
$$1=\frac{x-y}{x-y}=\frac{x-c}{x-y}+\frac{c-y}{x-y}$$
and with the calculations (using the triangle inequality in the "$\leq$" step)
\begin{align*}
\left|\frac{f(x)-f(y)}{x-y}-f'(c)\right| &=\left|\frac{f(x)-f(c)}{x-y}+\frac{f(c)-f(y)}{x-y}-\left(\frac{x-c}{x-y}+\frac{c-y}{x-y}\right)f'(c)\right|\\
&\leq\left|\frac{f(x)-f(c)}{x-y}-\frac{x-c}{x-y}f'(c)\right|+\left|\frac{f(c)-f(y)}{x-y}-\frac{c-y}{x-y}f'(c)\right|\\
&=\left(\frac{x-c}{x-y}\right)\left|\frac{f(x)-f(c)}{x-c}-f'(c)\right|+\left(\frac{c-y}{x-y}\right)\left|\frac{f(c)-f(y)}{c-y}-f'(c)\right|
\end{align*}
Addendum:
I had missed you asking about the second inequality. Here it goes:
By the fact that $f:I=[a,b]\to \mathbb{R}$ is differentiable at $c\in I$, we can find, for any arbitrarily fixed $\epsilon > 0$
a $\delta:=\delta(\epsilon)>0$ such for all $x\in I$ with $|x-c|<\delta$, we have $$\left|\frac{f(x)-f(c)}{x-c}-f'(c)\right|<\epsilon.$$
Start from the last expression above and assume $|x-c|<\delta$ and $|c-y|<\delta$. We get
$$\left(\frac{x-c}{x-y}\right)\left|\frac{f(x)-f(c)}{x-c}-f'(c)\right|+\left(\frac{c-y}{x-y}\right)\left|\frac{f(c)-f(y)}{c-y}-f'(c)\right| \\< 
\left(\frac{x-c}{x-y}\right)\epsilon + \left(\frac{c-y}{x-y}\right)\epsilon =\left(\frac{x-c}{x-y}+\frac{c-y}{x-y}\right)\epsilon = \epsilon
$$
