$\sqrt{\cdot}$ on $[0,\infty)$ is uniformly continuous and exactly $1/2$-Hölder, and is not Lipschitz

Let $X := [0,\infty) \subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) with the metric $d(x,y) = |x-y|$ inherited from $\mathbb{R}$ (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 $g : X \to X$ be $g(x) := \sqrt{x} = x^{1/2}$ (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$; the positives are $\{x^2 : x \neq 0\}$, Rational powers $a^r$ of a positive base). Then:

1. $\big|\sqrt{x} - \sqrt{y}\big| \le |x-y|^{1/2}$ for all $x,y \ge 0$, so $g$ is $1/2$-Hölder with constant $1$ (Lipschitz map, $\alpha$-Hölder map for rational $0 < \alpha \le 1$, and contraction).
2. $g$ is uniformly continuous (Uniform continuity of a map of metric spaces: one $\delta$ serving every point).
3. The constant $1$ cannot be improved: every $1/2$-Hölder constant $C$ for $g$ satisfies $C \ge 1$.
4. For every rational $\alpha$ with $1/2 < \alpha \le 1$, the map $g$ is not $\alpha$-Hölder. In particular, at $\alpha = 1$, $g$ is not Lipschitz.

So $1/2$ is exactly the Hölder exponent of the square root, and the example separates "Hölder" from "Lipschitz" inside Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent.

• Where the failure is located. The obstruction in claims 3 and 4 sits at $0$: the inequality $t^{1/2} \le C\,t^{\alpha}$ is comfortable for large $t$ and impossible for small $t$ once $\alpha > 1/2$, because $t^{1/2}$ then dominates $t^{\alpha}$. Away from $0$ the square root is perfectly Lipschitz: on $[a,\infty)$ with $a > 0$ one has $\big|\sqrt x - \sqrt y\big| = |x-y|/(\sqrt x + \sqrt y) \le |x-y|/(2\sqrt a)$, since $\sqrt x + \sqrt y \ge 2\sqrt a > 0$ and $(\sqrt x - \sqrt y)(\sqrt x + \sqrt y) = x - y$.
• The exponent is rational, and that is not a restriction here. At this page's position in the reading order, Rational powers $a^r$ of a positive base is the available exponent construction. Both $1/2$ and every exponent used above are rational, so claim 4 is intentionally local to rational $\alpha$; real powers are introduced later in Real powers for positive bases, with the zero-base positive-exponent convention ↗.
• What this example is for. It is one of the two witnesses named in Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent: it shows that "Hölder" is strictly weaker than "Lipschitz", so the implication proved there from Hölder to uniform continuity is not a detour through the Lipschitz condition. The other witness, separating continuity from uniform continuity, is $x \mapsto 1/x$ is continuous on $(0,1)$ and sends the Cauchy sequence $(1/(k+2))_{k \ge 0}$ to an unbounded one.
• Claim 1 is sharp in a second sense as well: equality holds whenever one of the two arguments is $0$, since $\big|\sqrt x - \sqrt 0\big| = \sqrt x = |x-0|^{1/2}$. So the estimate is attained and not merely approached.