If $f$ is $\mathscr{C}^1$, then $f(x) - f(y) = \int_0^1 Df(y + t(x-y)).(x-y) dt$, by the fundamental theorem of calculus.
Hence, $$\begin{aligned} \| f(x) - f(y) \| &\le& &\int_0^1 \|Df(y+t(x-y)).(x-y) \| dt& \\ &\le& &\left( \int_0^1 \| Df( y + t(x-y) )\| dt \right) \| x-y \|& \le \sup_{z \in \mathbb{R}^n} \, \|Df(z) \| \; \| x-y \| \end{aligned}$$
If $\sup_{z \in \mathbb{R}^n} \, \| Df(z) \| = C$ is finite, we get $\| f(x) - f(y) \| \le C \|x - y \|$ for all $x,y$.
Reciprocally, suppose that your function is $\mathscr{C}^1$ and that it's globally Lipschitz, with constant $C$.
Then, for all $x \in \mathbb{R}^n$, and all $h \in \mathbb{R}^n$, we know that $$Df(x).h = \lim_{t \to 0} \frac{f(x+th) - f(x)}{t}$$
But, by assumption, $\| f(x +th) - f(x) \| \le C \|th \| = C |t| \|h\|$, and we finally get $\|Df(x).h \| \le C \|h \|$ for all $h$, which by definition implies $\| Df(x) \| \le C$. Hence the total derivative is bounded all over $\mathbb{R}^n$.
All of this works also on an open set of $\mathbb{R}^n$, instead of the whole space.
Remark also that you don't need to assume $f$ to be $\mathscr{C}^1$, but only differentiable. The second part of my proof works as well, and for the first part, instead of applying fundamental theorem of calculus, you can use the mean value theorem.
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.
Best Answer
The function $f(x) = x \cdot \sin \frac{1}{x} \; (x > 0)$, $f(0) = 0$ is a counterexample.