Is a function Lipschitz if and only if its derivative is bounded? Is the following statement true?

Let $f: \mathbb{R}\to\mathbb{R}$ be continuous and differentiable.
$f$ Lipschitz $\leftrightarrow \exists M:\forall x\in\mathbb{R}\ |f'(x)|\leq M$

If $f'$ is bounded, it is Lipschitz, that's obvious.
Does that work the other way around?
Let $f$ be $M$-Lipschitz, that is to say $\forall x_1, x_2\in\mathbb{R},\ |f(x_1) - f(x_2)| \leq M|x_1 - x_2|$, where $M$ is independent of $x_1, x_2$.
Let $x_1 <x_2$ be arbitrary.
$f$ is continuous, so by the mean value theorem,
there exists $c\in\mathbb{R}, x_1 < c < x_2$, such that $f'(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \Rightarrow |f'(c)| = |\frac{f(x_2) - f(x_1)}{x_2-x_1}| \leq M$.
Does this hold? Can you, using the mean value theorem, "reach" every point in the derivative?
Also, another question: If $f$ is Lipschitz, is it necessarily differentiable?
Thanks!
 A: On request, here is a multidimensional version of Three.One.Four's answer.
\begin{align*}
\|\nabla f(x)\| 
&= \sup_{v} \frac{\|\nabla f(x)v\|}{\|v\|}
=\sup_{v} \lim_{\lambda\to0}\frac{|\lambda|\|\nabla f(x)v\|}{|\lambda|\|v\|}
=\sup_{v} \lim_{\lambda \to0}\frac{\|\nabla f(x)(\lambda v)\|}{\|\lambda v\|}\\
&\le \sup_{v} \lim_{\lambda \to0}
\frac{ \|f(x+\lambda v) - f(x) -\nabla f(x)(\lambda v)\|}{\|\lambda v\|}
+ \frac{ \|f(x+\lambda v) - f(x) \|}{\|\lambda v\|}\\
&\le \sup_{v} \lim_{\lambda \to0}
\frac{ \|f(x+\lambda v) - f(x) -\nabla f(x)(\lambda v)\|}{\|\lambda v\|}
+ \frac{ M\|(x+\lambda v) - (x) \|}{\|\lambda v\|}\\
&= \sup_{v} \underbrace{\lim_{\lambda \to0}
\frac{ \|f(x+\lambda v) - f(x) -\nabla f(x)(\lambda v)\|}{\|\lambda v\|}}_{=0}
+ M\\
&\le M
\end{align*}
Note that we simply used the definition of the multidimensional derivative here.
A: Assume that $f$ is a M-Lipschitz function. So $|f(x+h)-f(x)|\leq M|h|, \quad \forall x, h \in \mathbb{R}.$
It is equivalent to $|\frac{f(x+h)-f(x)}{h}|< M$. By taking que limit, $|f'(x)|\leq M$.
For the converse, use the mean value theorem. Let $x,y \in \mathbb{R}$, there exists $c\in \mathbb {R}$ so that $f(x)-f(y)=(x-y)f'(c)$ (you have to assume that $f$ is differentiable) and now use the fact that $|f'(c)|\leq M$.
As the previous poster said, Rademacher's theorem says that every Lipschitz function is almost everywhere differentiable. 
A simple example of non differentiable Lipschitz function is the absolute value. 
A: You don't have to use the mean value theorem. Just use the definition of the derivative:
$$ | f'(x) | = \lim_{h \to 0} \frac{|f(x+h)-f(x)| }{|h|} \leq \lim_{h \to 0} \frac{M |x+h -x|}{|h|} = M.$$
And no, Lipschitz functions don't have to be differentiable, e.g. the absolute value $| \cdot |$ is Lipschitz. Lipschitz functions are differentiable almost everywhere though.
