Let $d$ be a complete metric for $X$. Let $f: X \to X$ be a function. Suppose there is a number $k$, with $0 < k < 1$, such that $d(f(x), f(y)) \leq kd(x, y)$ for all $x, y \in X$. Then $f$ is continuous and has exactly one fixed point.
To show that $f$ is continuous, it felt like a good idea to use sequential continuity. So we take a convergent sequence $x_n$ with limit $x$. Since $X$ iss complete, we know that $x_n$ is a Cauchy sequence. So for every $\varepsilon/k > 0$ we have some $N \in \mathbb N$ such that for every $n, m \geq N$ we have $d(x_n, x_m) < \varepsilon/k$. But, if $d(x_n, x_m) < \varepsilon/k$, then $d(f(x_n), f(x_m)) < \varepsilon$. So $f(x_n)$ is a Cauchy sequence which converges to $f(x)$. Hence $f$ is continuous.
My problem is how to show that $f$ has exactly one fixed point. What I want to do is assume that $f$ has no fixed points, and then assume that $f$ has at least two, and obtain contradictions. However, I'm not sure how to proceed after assuming $f$ has no fixed points.