This is Exercise $1.5.4$ in Introduction to Dynamical Systems by Michael Brin and Garrett Stuck.
Let $f: \Bbb R \to \Bbb R$ be a $C^1$ map, and $p$ be a fixed point. Show that if $|f'(p)|<1$, then $p$ is attracting, and if $|f'(p)| > 1$, then $p$ is repelling.
I'll recall the definition of an attracting fixed point. $p$ is an attracting fixed point if $f(p) = p$, and there exists a neighborhood $U \ni p$ such that $\overline{U}$ is compact, $f(\overline{U}) \subset U$, and $\bigcap_{n\ge 0} f^n(U) = \{p\}$.
My work: Assume $|f'(p)| < 1$. There exists $\delta > 0$ such that $|f'(p)| < 1- \delta < 1$. By continuity of $f'$, there exists $\epsilon > 0$ such that $x\in (p-\epsilon,p+\epsilon)$ implies $|f'(x)| < 1-\delta$. Define $U := (p-\epsilon, p+\epsilon)$. For any $x\in U$, $$|f(x) - f(p)| < (1-\delta)|x-p|$$ by the Mean Value Theorem. So, $$|f(x)-p| < (1-\delta)\epsilon < \epsilon.$$ Now, if $x\in \overline{U}$, there exists a sequence $\{x_n\} \subset U$ converging to $x$. Taking $\lim_{n\to\infty}$ in $$|f(x_n)-p| < (1-\delta)\epsilon,$$ we get $$|f(x)-p| \leq (1-\delta)\epsilon < \epsilon.$$ It follows that $f(\overline{U}) \subset U$.
I want to remark that $\lim_{n\to\infty} f^n(x) = p$ for all $x\in U$, in case that is helpful. This is because we have $$|f^n(x) - p| = |f^n(x) - f^n(p)| < (1-\delta)^n \epsilon$$ using a Mean-Value theorem argument like above. Taking $\lim_{n\to\infty}$ gives $\lim_{n\to\infty}|f^n(x) - p|=0$.
It remains to show $\bigcap_{n\ge 0} f^n(U) = \{p\}$. Consider $y\in \bigcap_{n\ge 0} f^n(U) = \{p\}$. There exists a sequence $\{y_n\} \subset U$ such that $y = f^n(y_n)$ for all $n\ge 0$. Taking limits, we see that $\{f^n(y_n)\}$ is convergent, and $y = \lim_{n\to\infty} f^n(y_n)$. This is where I'm stuck, and I hope the solution up to this point is correct. Since I am (also) asking for a (partial) proof-verification, this should not be a duplicate of other posts.
Thanks for your help!