[proofplan] We prove uniqueness by comparing two arbitrary fixed points. The contraction inequality turns the equality $f(x)=x$ and $f(y)=y$ into an inequality of the form $d(x,y) \le c\,d(x,y)$ with $c<1$. Rearranging forces $d(x,y)=0$, and the definiteness axiom of the metric then gives $x=y$. [/proofplan] [step:Compare two arbitrary fixed points using the contraction inequality] Let $x,y \in X$ be fixed points of $f$, meaning that $f(x)=x$ and $f(y)=y$. Since $f$ is a contraction, there exists a constant $c \in [0,1)$ such that \begin{align*} d(f(a),f(b)) \le c\,d(a,b) \end{align*} for all $a,b \in X$. Applying this inequality with $a=x$ and $b=y$, and then using the fixed point identities, gives \begin{align*} d(x,y)=d(f(x),f(y)) \le c\,d(x,y). \end{align*} [guided] Let $x,y \in X$ be two fixed points of $f$. By the definition of fixed point, this means \begin{align*} f(x)=x \end{align*} and \begin{align*} f(y)=y. \end{align*} The goal is to prove that these two points are actually the same point. The only quantitative information about $f$ is that it is a contraction. Thus there exists a constant $c \in [0,1)$ such that, for every pair of points $a,b \in X$, \begin{align*} d(f(a),f(b)) \le c\,d(a,b). \end{align*} We apply this universal inequality to the particular pair $a=x$ and $b=y$. This gives \begin{align*} d(f(x),f(y)) \le c\,d(x,y). \end{align*} Because $x$ and $y$ are fixed points, the left-hand side is exactly the distance between $x$ and $y$: \begin{align*} d(f(x),f(y))=d(x,y). \end{align*} Substituting this identity into the contraction inequality yields \begin{align*} d(x,y) \le c\,d(x,y). \end{align*} This is the decisive inequality: a nonnegative number is at most a strict fraction of itself. [/guided] [/step] [step:Use the strict inequality $c<1$ to force the distance to vanish] From \begin{align*} d(x,y) \le c\,d(x,y) \end{align*} we subtract $c\,d(x,y)$ from both sides and obtain \begin{align*} (1-c)d(x,y) \le 0. \end{align*} Since $c \in [0,1)$, we have $1-c>0$. Since $d$ is a metric, $d(x,y) \ge 0$. Hence the product $(1-c)d(x,y)$ is both nonnegative and at most $0$, so \begin{align*} (1-c)d(x,y)=0. \end{align*} Dividing by the positive number $1-c$ gives \begin{align*} d(x,y)=0. \end{align*} [/step] [step:Apply definiteness of the metric to identify the fixed points] The definiteness axiom for the metric $d$ states that $d(u,v)=0$ for $u,v \in X$ implies $u=v$. Applying this axiom to the points $x,y \in X$ and using $d(x,y)=0$, we conclude that \begin{align*} x=y. \end{align*} Thus any two fixed points of $f$ coincide, so $f$ has at most one fixed point. [/step]