Let $f\in C^1(\mathbb{R})$ and let $x_0\in P$ where $P$ is a perfect set.I want proof that $\forall \varepsilon>0$$\exists\delta>0$ with $\lvert f(x)-f(y)-f'(y)(x-y) \rvert<\varepsilon\lvert x-y\rvert$$\forall x,y\in P$ ,$\lvert x-x_0\rvert<\delta$,$\lvert y-x_0\rvert<\delta$.
My attempt: let $\varepsilon>0$.Because $F'$ is continous in $x_0$, exists $\delta>0$ with $\lvert f'(x)-f'(x_0)\rvert<\dfrac{\varepsilon}{2}$ for all $x\in\mathbb{R}$ , $\lvert x-x_0\rvert<\delta$.
Let now $x,y\in P$, $x\neq y$ with $\lvert x-x_0\rvert<\delta $ and $\lvert y-x_0\rvert<\delta$.For MVT theorem exists $x'\in (x,y)\cup (y,x)$ with $f'(x')=\dfrac{f(y)-f(x)}{y-x}$.
Then $\lvert f(x)-f(y)-f'(y)(x-y)\rvert=\biggl\lvert \dfrac{f(x)-f(y)}{x-y}-f'(y)\biggr\rvert\lvert x-y\rvert=\lvert f'(x')-f'(y)\rvert\lvert x-y\rvert<[ \lvert f'(x')-f'(x_0)\rvert+\lvert f'(x_0)-f'(y)\rvert] \lvert x-y\rvert$.
I control $\lvert f'(x_0)-f'(y)\rvert<\dfrac{\varepsilon}{2}$ because $\lvert y-x_0\rvert<\delta$.
How i can control instead $\lvert f'(x')-f'(x_0)\rvert$?I have to distinguish position of $x,y,x_0,x'$ to evaluate $\lvert x'-x_0\rvert$ or the hypothesis that $P$ is perfect is relevant?
Thanks in advance.