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

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.