Prove Taylor expansion with mean value theorem

Equation (8) is certainly not true. A simple counterexample is $f(t)=t^3, x=0, \Delta=1,$ because we get $f'(\xi_1)=3\xi_1^2=f(1)-f(0)=1, \xi_1={1\over \sqrt3}$ and hence $\Delta_2=\xi_1={1\over \sqrt3}\neq {\Delta \over 2}.$ Also Taylor's Theorem doesn't say that an infinitely differentiable function necessarily coincides with its Taylor series. All Taylor's Theorem does is give a measure for the error made by estimating a function with its $n^{th}$ Taylor polynomial.

If you want to learn about the connection between the Taylor expansion and the Mean Value Theorem I suggest:
http://en.wikipedia.org/wiki/Taylor%27s_theorem
http://www.proofwiki.org/wiki/Taylor%27s_Theorem/One_Variable

A non-rigorous approach can be writing the power series and finding the coefficients through reasoning...

$$ f(x) = a_0 + a_1(x-a) + a_2(x-a)^2 + \dots + a_n(x-a)^n + \dots $$

from there $a_0=f(a)$, $a_1= f'(a)$ and successive terms can be found by differentiating both sides further.