$\sqrt{x}$ isn't Lipschitz function A function f such that 
$$
|f(x)-f(y)| \leq C|x-y|
$$
for all $x$ and $y$, where $C$ is a constant independent of $x$ and $y$, is called a Lipschitz function
show that $f(x)=\sqrt{x}\hspace{3mm} \forall x \in \mathbb{R_{+}}$ isn't Lipschitz function
Indeed,  there is no such constant C where 
$$
|\sqrt{x}-\sqrt{y}| \leq C|x-y| \hspace{4mm} \forall x,y \in \mathbb{R_{+}}
$$
we have only that inequality
$$
|\sqrt{x}-\sqrt{y}|\leq |\sqrt{x}|+|\sqrt{y}|
$$
Am i right ?
remark for @Vintarel  i plot it i don't know  graphically "Lipschitz" mean? 
what is the big deal in the graph of the square-root function 

in wikipedia they said 
Continuous functions that are not (globally) Lipschitz continuous The function f(x) = $\sqrt{x}$ defined on [0, 1] is not Lipschitz continuous. This function becomes infinitely steep as x approaches 0 since its derivative becomes infinite. However, it is uniformly continuous as well as Hölder continuous of class $C^{0,\alpha}$, α for $α ≤ 1/2$.
Reference
1] could someone explain to me this by math and not by words, please ??
2] what does "Lipschitz" mean graphically?

Showing $\sqrt {x}$ is not Lipschitz .
We prove this by contradiction method.
Suppose if possible $\sqrt x$ is Lipschitz function. Then there exists $C>0$ such that
$|√x-√y|<C|x-y|$ for all $x,y\in \mathbb R^{+}$.
$\\$
Take $x=(\frac{1}{1+C})^2$ and $y=0$.
Then $|\frac{1}{1+C}|<C((\frac{1}{1+C})^2)$
$\Rightarrow C+1\leq C$ which is not possible.
So $√x$ is not Lipschitz function.
