$\sqrt{x}$ isn't Lipschitz function

Hint: why is it not possible to find a $C$ such that $$ |\sqrt{x} - \sqrt{0}|\leq C|x-0| $$ For all $x \geq 0$?

As a general rule: Note that a differentiable function will necessarily be Lipschitz on any interval on which its derivative is bounded.

In response to the wikipedia excerpt: "This function becomes infinitely steep as $x$ approaches $0$" is another way of saying that $f'(x) \to \infty$ as $x \to 0$. If you look at slope of the tangent line at each $x$ as $x$ gets closer to $0$, those tangent lines become steeper and steeper, approaching a vertical tangent at $x = 0$.

"Graphically", we can say that a differentiable function will be Lipschitz (if and) only if it never has a vertical tangent line.

Some functions that are not Lipschitz due to an unbounded derivative: $$ f(x) = x^{1/3}\\ f(x) = x^{1/n},\quad n = 2,3,4,5,\dots $$ A more subtle example: $$ f(x) = x^2,\quad x \in \mathbb{R}\\ f(x) = \sin(x^2), \quad x \in \mathbb{R} $$ Note in these cases that although $f'(x)$ is continuous, there is no upper bound for $f'(x)$ over the domain of interest.

You have $${\sqrt{1/n} - \sqrt{0}\over{1/n - 0}} = {1/\sqrt{n}\over {1\over n}} = \sqrt{n}.$$ This ratio can be made as large as you like by choosing $n$ large. Therefore the square-root function fails to be Lipschitz.

Suppose that $\sqrt{x}$ is a Lipschitz function, then there exists $C$ such that

$$\Big|\frac{\sqrt{y}-\sqrt{x}}{y-x}\Big| \le C$$

Now, Let $y=2x$, so

$$(\sqrt{2}-1)x^{-\frac{1}{2}}\le C$$

Letting $x→0$ gives a contradiction.