real analysis - Prove that if $f$ is differentiable at $x_0$, then $f$ satisfies Lipschitz condition at $x_0$

A function $f:(a,b)\rightarrow \mathbb{R}$ satisfies a Lipschitz condition at $x_0 \in (a,b)$ if $\exists M >0$ and $\mu >0$ for which the following is true:

if $y\in(a,b)$ and $| x_0 -y|<\mu$, then $|f(x_0) -f(y)\leq M|x_0-y|$

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ROAD MAP TO PROBLEM (provided by professor)

a) Why must there exists a $\delta >0$ for which the following is true:

if $y\in(a,b)$ and $0<| x_0 -y|<\delta$, then $| \frac{f(x_0) -f(y)}{x_0 -y} -f'(x_0)|<1$

b) show (briefly) that for all $y \in (a,b)$
$$|f(x_0)-f(y)|=|\frac{f(x_0) -f(y)}{x_0 -y}-f'(x_0)+f'(x_0)||x_0 -y|$$

c) Prove that there exists a number $M>0$ for which, if $y\in (a,b)$ and $0<|x_0-y|<\delta$, then $$|f(x_0)-f(y)|\leq M|x_0 -y|$$ Hint: $M$ will depend on the value of f at $x_0$

d) Explain ( briefly), why, if $y=x_0$, then $|f(x_0) - f(y)| \leq M|x_0 -y|$

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Given that f is differentiable thus, f is continuous, therefore I can somewhat see how they got to (a), however after (a) I'm confused as to how to look at the problem.

Answer

Assume that $f'(x_0) \neq 0$

For $\epsilon=1$ exists $\delta>0$ such that $\frac{|f(x)-f(x_0)|}{|x-x_0|} <1+|f'(x_0)|,\forall x \in(x_0-\delta,x_0+\delta)$

So for $M=1+|f'(x_0)|$ you have the conclusion.

We used that if: $\lim_{x \to x_0}g(x)=l$ then $\lim_{x \to x_0}|g(x)|=|l|$