The square root: uniformly continuous, not Lipschitz

1. By symmetry assume $x\ge y\ge0$. We want $\sqrt x\le\sqrt y+\sqrt{x-y}$. Both sides are non-negative, so it is equivalent to the inequality of squares:
$$x\le y+(x-y)+2\sqrt{y(x-y)}=x+2\sqrt{y(x-y)},$$
which is true. Hence $\sqrt x-\sqrt y\le\sqrt{x-y}$, and the left side is $\ge0$.

2. Let $\varepsilon>0$ and $\delta=\varepsilon^2$. If $|x-y|<\delta$, then $|\sqrt x-\sqrt y|\le\sqrt{|x-y|}<\sqrt\delta=\varepsilon$. The choice of $\delta$ does not depend on the point.

3. If $\sqrt x$ were $K$-Lipschitz on $[0,1]$, then taking $y=0$ gives $\sqrt x\le Kx$, i.e. $\frac1{\sqrt x}\le K$ for all $x\in(0,1]$. But $\frac1{\sqrt x}\to+\infty$ as $x\to0^+$. Contradiction.

Remark. So Lipschitz is strictly stronger than uniformly continuous. The problem is the vertical tangent of the graph of $\sqrt x$ at $0$: the slope is unbounded, but the function still moves slowly enough that $\delta=\varepsilon^2$ works.