If $f$ is differentiable on $[a,b]$, then it is also Lipschitz on it. He guys, I am trying to show that a differentiable function defined on a closed interval is also Lipschitz on it. I managed to weave the below proof, but I have a feeling that it may be just a tad too general for this purpose:
Theorem. If $f$ is differentiable on $[a,b]$, then it is also Lipschitz on it.
Recall that $f:A\to\mathbb{R}$ is Lipschitz on $A$ if there exists an $M>0$ such that$$\left|\frac{f(x)-f(y)}{x-y}\right|\leq M$$for all $x,y\in A$.
Proof. Let $f$ be differentiable on $[a,b]$. Because $f$ is continuous and $[a,b]$ is compact, by the Extreme Value Theorem, it follows that $f$ attains a maximum value $M$. Moreover, since $f$ is differentiable on $[a,b]$,$$f'(y)=\lim_{x\to y}\left|\frac{f(x)-f(y)}{x-y}\right|\leq M,$$for all $x,y\in[a,b]$, as required. $\square$
What do you guys think?
Edit: What if we were to add that $f'$ is also continuous on $[a,b]$?

It is not true in general. Consider function
$$
f(x)=
\begin{cases}
x^2 \sin\frac{1}{x^2}\qquad x\neq 0\\
0\qquad\quad\qquad x=0
\end{cases}
$$
It is differentiable on $[0,1]$ and
$$
f'(x)=
\begin{cases}
2 x \sin\frac{1}{x^2}-\frac{2}{x} \cos\frac{1}{x^2},  &x\neq 0\\
0,  &x=0
\end{cases}
$$ 
But this derivative is unbounded, since
$$
\lim\limits_{n\to\infty} f'\left(\frac{1}{\sqrt{ {\pi} +2\pi n}}\right)=+\infty
$$
On the other hand if require $f'\in C([0,1])$ then by Weierstrass theorem there exist 
$$
M=\sup\limits_{t\in [0,1]}|f'(t)|<+\infty.
$$
This $M$ is a Lipschitz constant you are looking for.
