Call $d(a,b) = \frac{f(a)-f(b)}{a-b}$ and $\delta(a,b) =$ the unique $\xi \in ]a;b[$ such that f'(\xi) = d(a,b).
For any $a \in \mathbb{R}$, the function $d_a : b \mapsto d(a,b)$, extended on $b=a$ with d_a(a) = f'(a) is continuous and injective. To prove the injectiveness, if you have $d(a,b) = d(a,c)$, then $d(a,b) = d(a,c) = d(b,c)$, and if you look at the bigger of the three intervals, you will find one value for $\xi$ in each of the smaller intervalls, which contradicts your hypothesis.
Therefore, for every $a$, $d_a$ is either strictly increasing or strictly decreasing. From this and from the continuity of $d$, you get that either they are all strictly increasing, either they are all strictly decreasing.
Without loss of generality, suppose they are increasing. Then you get that f' is also strictly increasing : if $a \lt b$, then f'(a) = d_a(a) \lt d_a(b) = d_b(a) \lt d_b(b) = f'(b).
From here, choose $x_0, x_1$ and build the sequence $\xi_n$ as you described. $\xi_n$ is a strictly decreasing sequence, bounded below by $x_0$ so it converges to some limit $\xi$. If $\xi \gt x_0$ then f'(\xi) = f'(\lim \xi_n) \le \lim f'(\xi_n) = \lim d(x_0, \xi_{n-1}) = d(x_0, \xi) = f'(\delta(x_0, \xi)) \lt f'(\xi), which is impossible. Thus $\lim \xi_n = x_0$.