$f(x)=x^2$ is not Lipschitz?

Consider the function $\ f:\mathbb{R}\to\mathbb{R}$ defined by $\ f(x)=x^2$ for all $x\in \mathbb{R}$.

I think that $f$ is NOT Lipschitz (meaning, there exists no constant $c\in[0,\infty)$ such that $|f(x)-f(y)|\leq c |x-y|$ for all $x,y\in\mathbb{R}$). Here is my proof:

Suppose that it IS Lipschitz. There there exists some constant $c\in[0,\infty)$ such that $|f(x)-f(y)|\leq c |x-y|$ for all $x,y\in\mathbb{R}$. Then:

$\implies\ \ \ \ |x^2-y^2|\leq c |x-y|$ for all $x,y\in\mathbb{R}$

$\implies\ \ \ \ |x-y| \cdot |x+y|\leq c |x-y|$ for all $x,y\in\mathbb{R}$

$\implies\ \ \ \ |x+y|\leq c$ for all $x,y\in\mathbb{R}$

But there is no real number $c$ with the property that $|x+y|\leq c$ for all $x,y\in\mathbb{R}$. This is a contradiction. $f$ is not Lipschitz.

Is this proof correct?

A shorter and more eloquent way to show it is not Lipschitz is fix $x$ at $0$, and consider a sequence tending to $\infty$, say, $\{n\}$, then $$|f(n) - f(0)|/|n - 0| = n^2/n = n,$$ which cannot be bounded by any fixed Lipschitz constant $c$.

Your proof is mostly correct but here is a shorter proof:

If $f$ were Lipschitz, then $f'$ would be bounded.