Prove that a differentiable function $f\colon I\to\mathbb{C}$ on an interval $I$ with bounded derivative is Lipschitz continuous, i.e. If $\lvert f'\rvert\leq L $ for some $L\in\mathbb{R}$, then for any $x_1,x_2\in I$ we have $$ \lvert f(x_1)-f(x_2)\rvert\leqslant L\lvert x_1-x_2\rvert. $$
Despite two little things, I think the proof should work as follows:
I think, we can choose some $c\in\mathbb{C}$ with $\lvert c\rvert =1$ such that $$ \lvert f(x_2)-f(x_1)\rvert = c\cdot (f(x_2)-f(x_1))~~(*) $$ since for $$ \frac{f(x_2)-f(x_1)}{\lvert f(x_2)-f(x_1)\rvert}=:v $$ we have $\lvert v\rvert =1$ and then we can define $c:=v^{-1}$.
Next, in order to have equation $(*)$, the LHS has to be the real part of the RHS, i.e. $$ \lvert f(x_2)-f(x_1)\rvert =\Re(c\cdot (f(x_2)-f(x_1)))=\varphi(x_2)-\varphi(x_1), $$ where $$ \varphi:=\Re(cf). $$
Now, since $f$ is differentiable on $I$, it is, in particular, continuous on $[x_1,x_2]$, hence $\varphi$ is also continuous on $[x_1,x_2]$.
I am not sure about the following question:
Do we also have that $\varphi$ is differentiable on $(x_1,x_2)$? (Q)
Assuming that we can answer question (Q) with YES, we could apply the mean value Theorem on $\varphi$, telling us that $$ \varphi(x_2)-\varphi(x_1)=(x_2-x_1)\varphi'(\xi) $$ for some $\xi\in (x_1,x_2)$.
By assumption, $\varphi'(\xi)=\Re(cf'(\xi))\leq\lvert cf'(\xi)\rvert\leq L$ and hence $$ \lvert f(x_2)-f(x_1)\rvert = \varphi(x_2)-\varphi(x_1)=(x_2-x_1)\varphi'(\xi)\leqslant L\lvert x_2-x_1\rvert. $$
I am also not completely sure if $$ \varphi' = \Re(cf') $$ is correct.
Despite the two things in the two yellow boxes, I am pretty sure the proof should work. It would be nice if you could give me some hints.