x^2 is not uniformly continuous on the real line

Counterexample

The function $f \colon \mathbb{R} \to \mathbb{R}$ with $f(x) = x^2$ is continuous but not uniformly continuous. For $n \ge 1$, let $x_n = n$ and $y_n = n + 1/n$. Then $|x_n - y_n| = 1/n \to 0$, but $$f(y_n) - f(x_n) = 2 + \frac{1}{n^2} > 2.$$ So no $\delta$ works for $\varepsilon = 2$.

Remarks

On each bounded interval $[-c, c]$, the function $x^2$ is Lipschitz with constant $2c$. So it is uniformly continuous there. The problem is that the slope $2x$ is not bounded on $\mathbb{R}$.

Explanations

L1 human

The graph of $x^2$ gets steeper and steeper as $x$ grows. So a fixed small change in $x$ gives larger and larger changes in $x^2$.