$f$ is injective because if $x\neq x'$ , $d(f(x),f(x'))=d(x,x')>0$ . We show it is surjective. Suppose not. Then, there is some $y\in X-f(X)$ . Since $f$ is continuous (Theorem 21.1) and $X$ is compact, $f(X)$ is compact, and, hence, closed. Therefore, there exists an $\epsilon$ -neighborhood of $y$ that does not intersect $f(X)$ . Now, using the hint, let $x_{1}=y$ , and $x_{n+1}=f(x_{n})$ . Then, $d(x_{1},x_{n})\ge\epsilon$ for all $n>1$ , because $x_{1}=y$ and $x_{n}\in f(X)$ . Further, for $n>m>1$ , $d(x_{n},x_{m})=d(f(x_{n-1}),f(x_{m-1}))$ $=d(x_{n-1},x_{m-1})$ $=\cdots$ $=d(x_{n-m+1},x_{1})$ $\ge\epsilon$ . Therefore, we have an infinite sequence of pairwise distinct points such that no subsequence converges (if there was a converging subsequence, it would be within $\tfrac{\epsilon}{2}$ -ball of some point starting from some index, and pairwise distances between the points of the subsequence starting from this index would be less than $\epsilon$ ). This means that $X$ is not sequentially compact, but $X$ is a compact metric space. This fact contradicts Theorem 28.2. Hence, there is no such point $y$ , and $f$ is surjective. Also, by Exercise 2 of §21, $f$ is a homeomorphism.