Continuous differentiability implies Lipschitz continuity

By the mean value theorem,

$$f(x) - f(y) = f'(\xi)(x-y)$$ for some $\xi \in (y,x)$. But since $f'$ is continuous and $[a,b]$ is compact, then $f'$ is bounded in that interval, say by $C$. Thus taking absolute values yields

$$\lvert f(x) - f(y)\rvert \le C \lvert x-y\rvert$$

Since $f'$ is continuous it is bounded on $[a,b]$. Choose $M\in \Bbb R$ such that $|f'(x)|\le M$ for all $x\in [a,b]$. If $x,y\in [a,b]$, then by the MVT there is $\xi\in [a,b]$ with $f(x)-f(y)=f'(\xi)(x-y)$, thus $|f(x)-f(y)|\le M|x-y|$. In fact you can show that a differentiable function on an open interval (not necessarily a bounded interval) is Lipschitz continuous if and only if it has a bounded derivative. This is because any Lipschitz constant gives a bound on the derivative and conversely any bound on the derivative gives a Lipschitz constant.

To your other question: Continuous derivative does not imply twice differentiable. In fact, we know that there are functions $f$ (such as the Weierstrass-function) which are continuous but not differentiable. Taking the antiderivative $F$ gives us a differentiable function with derivative $f$ (by the Fundamental Theorem of Calculus), so it has a continuous derivative, but as $f$ is not differentiable, $F$ is not twice differentiable.