Proof of The Absolute Value on the Real Line is 1-Lipschitz and Continuous

The Lipschitz bound. Let $x,y\in\mathbb{R}$. By Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (x),

$$d_{\mathbb{R}}\bigl(\alpha(x),\alpha(y)\bigr)=\bigl||x|-|y|\bigr|\le|x-y|=1\cdot d_{\mathbb{R}}(x,y),$$

so $\alpha$ is Lipschitz with constant $1$, which is (i).

Continuity. Since $d_{\mathbb{R}}$ is a metric on $\mathbb{R}$ (The Absolute Value Metric on the Real Numbers (i)), the Lipschitz bound just proved places $\alpha$ among the Lipschitz maps from the metric space $(\mathbb{R},d_{\mathbb{R}})$ to itself, and such a map is continuous (A Lipschitz Map is Uniformly Continuous and Continuous (ii)). Continuity with respect to $d_{\mathbb{R}}$ is continuity on $\mathbb{R}$ in the sense of the real line (The Real Line: Standing Notation and Background for Calculus on Intervals (vi)), which proves (ii).