Proof to show function f satisfies Lipschitz condition when derivatives f' exist and are continuous
Solution 1:
1) Continuous function $g$ in a compact interval $[a,b]$ implies $g$ bounded in $[a,b]$. (This part I make a bit more general than yours)
Consider the set $X=\{x\in[a,b]: g|[a,x] \text{ is bounded }\}$. Show that $a\in X$ and $\sup X\in X$ then $b=\sup X$ so $b\in X$.
2) Bounded derivative implies Lipschitz.
If $|f'(x)|\leq K$ for all $x\in[a,b]$, in particular for any $x,y\in[a,b]$ such that $x\neq y$ we have
$$\left|\dfrac{f(x)-f(y)}{x-y}\right|=|f'(c)|\leq K$$
for some $c\in[x,y]$ by MVT. Then $|f(x)-f(y)|\leq K|x-y|$.