This is Exercise $1.5.2$ in Introduction to Dynamical Systems by Michael Brin and Garrett Stuck.
Let $X$ be a locally compact metric space and $f:X\to X$ be a continuous map. We say that a fixed point $p$ is repelling if $f(p) = p$, and there exists a neighborhood $U$ of $p$ such that $\overline{U} \subset f(U)$, and $\bigcap_{n\ge 0} f^{-n}(U) = \{p\}$. Show that a repelling fixed point is an isolated fixed point, i.e., there exists a neighborhood $V$ of $p$ that contains no other fixed points.
If this is not the case, then for every $\epsilon > 0$, we can find a point $x \in X$ different from $p$, such that $d(x,p) < \epsilon$, and $f(x) = x$. Thus, we can find a sequence of fixed points $\{x_n\}$ converging to $p$, such that $d(x_n,p) < 2^{-n}$ for all $n\ge 1$. As $x_n \to p$, there exists $N\in \mathbb N$ such that $x_n \in U$ for all $n\ge N$. I'm not sure where to go from here. Thanks for any hints!
$f^{-1}(U)$ here means the inverse image of $U$ under $f$: i.e., $\{x : f(x) \in U\}$. And $f^{-n}$ means the $n$-th iterate of the inverse image operation. If you take $V = U$, then $V$ contains no fixed points other than $p$: if $q \in U$ were some other fixed point, then you'd have that $q \in f^{-n}(U)$ for every $n$ (because $f(q) = q$), so that $q \in \bigcap_{n \ge 0}f^{-n}(U)$ contradicting the assumption that $ \bigcap_{n \ge 0}f^{-n}(U) = \{p\}$