x² is not uniformly continuous on the real line

Negation. $f$ is *not* uniformly continuous if there is an $\varepsilon_0>0$ such that for every $\delta>0$ one can find $x,y$ with $|x-y|<\delta$ and $|f(x)-f(y)|\ge\varepsilon_0$.

On $\mathbb{R}$. Take $x_n=n+\frac1n$ and $y_n=n$. Then $|x_n-y_n|=\frac1n\to0$, but
$$x_n^2-y_n^2=2+\frac1{n^2}\ge2.$$
So $\varepsilon_0=2$ works: however small $\delta$ is, for $n>\frac1\delta$ the points $x_n,y_n$ are $\delta$-close with values at least $2$ apart.

On $[0,5]$. For $x,y\in[0,5]$, $|x^2-y^2|=|x-y|\,|x+y|\le10|x-y|$. So $f$ is $10$-Lipschitz on $[0,5]$, hence uniformly continuous.

Moral. The faster the graph climbs, the smaller $\delta$ must be. On the whole line the slope $2x$ is unbounded, so no single $\delta$ can do. On a bounded interval it is bounded.