$x \mapsto x/2$ maps $(0,1]$ into itself, is a $1/2$-contraction, and has no fixed point

Refuted claim: the completeness hypothesis in Banach's fixed point theorem (A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point) may be dropped; a contraction of a nonempty metric space into itself always has a fixed point.

Let $X := (0,1] \subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) carry the metric $d(x,y) := |x-y|$ inherited from the real line (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset), and let $f : X \to X$ be $f(x) := x/2$. Then $X$ is nonempty, $f$ maps $X$ into itself and is a contraction with constant $1/2$ (Lipschitz map, $\alpha$-Hölder map for rational $0 < \alpha \le 1$, and contraction), and $f$ has no fixed point in $X$. The single hypothesis of A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point that fails is completeness (Complete metric space: every Cauchy sequence converges in the space), and it does fail.

• The missing point is exactly the fixed point. In $\mathbb{R}$ the map $x \mapsto x/2$ has the fixed point $0$, and the iterates $x_n = x_0/2^{\,n}$ from any starting point in $(0,1]$ converge to $0$. They are Cauchy in $(0,1]$, as Banach's proof guarantees, and the space simply has nowhere to put their limit. So the theorem's proof runs correctly up to the last step, and completeness is precisely what that last step needs.
• Closedness would fix it. $[0,1]$ is closed in $\mathbb{R}$, hence complete (A subspace of a complete metric space is complete iff it is closed, and a complete subspace of any metric space is closed, $\mathbb{R}$ and $\mathbb{R}^n$ for $n \ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$), and the same map on it has the fixed point $0$. This is the usual way the hypothesis is met in practice.
• Every other hypothesis is present, which is what makes this a clean counterexample rather than a curiosity: $X$ is nonempty, $f$ maps $X$ into $X$, and the contraction constant $1/2$ is explicit and uniform.