Why $\delta(X_n) \rightarrow 0$?
I cannot understand the following proof in "metric spaces & topological spaces" written by Wataru Takahashi.
Is the following proof correct or not?
Theorem:
Let $(X, d)$ be a complete metric space. Lef $f$ be a map from $X$ to $X$ such that $d(f(x), f(y)) \leq r d(x, y)$ for all $x, y \in X$ for some $r \in [0, 1)$.
Show that there exists a unique $x_0 \in X$ such that $f(x_0) = x_0$.
Proof:
Let $X_0 = X, X_1 = \overline{f(X)}, \cdots, X_n = \overline{f^n(X)}$.
For all $n, X_n$ is closed.
$X_n \neq \emptyset$.
$X_0 \supset X_1 \supset \cdots \supset X_n \supset \cdots$.
$d(f^n(x), f^n(y)) \leq r^n d(x, y)$ for all $x, y \in X$. So, $\delta(X_n) \rightarrow 0$ ($\delta(X_n)$ is the diameter of $X_n$)
By Cantor's Theorem,
$$\bigcap_n X_n = \{x_0\}$$ for some $x_0 \in X$.
$$f(\{x_0\}) = f(\bigcap_n X_n) \subset \bigcap_n f(X_n) \subset \bigcap_n X_{n+1} =\{x_0\}.$$
So, $f(x_0) = x_0$.
If $x_1 \neq x_0$ satisfies $f(x_1) = x_1$, then $d(x_0, x_1) = d(f(x_0), f(x_1)) \leq r d(x_0, x_1)$.This implies $1 \leq r$, but $0 \leq r < 1$. Contradiciton.