Let $f$ be a real function such that $f$ is differentiable and $|f'(c)| \leq M<1$. Let $a_1$ a real number and define $a_{n+1} = f(a_n)$. Then $(a_n)$ converges.
My attempt:
By the mean theorem, we have that $|a_{n+1} - a_n| = |f(a_n) - f(a_{n-1})| \leq M|a_n - a_{n-1}|$.
Then, $|a_{n+1} - a_n| \leq M^{n-1}|a_2 - a_1|$.
As $M < 1$, we know that $M^{n-1}$ converges to $0$.
You are almost there! What you need is to prove that the sequence $\{a_n\}$ is a Cauchy sequence. We can prove this applying what you just proved and the triangle inequality, indeed for any $p \ge 1$ $$|a_{n + p} - a_n| \le \sum_{i = n}^{n + p - 1}|a_{i + 1} - a_i| \le \sum_{i = n}^{n + p - 1}M^{i - 1}|a_2 - a_1|\le |a_2 - a_1|\sum_{i = n}^{\infty}M^{i - 1} \to 0\quad \text{as}\ n \to \infty.$$ The last limit follows from the fact that the term on the right hand side is the tail of a convergent (geometric) series.
