So we assume that $f$ is differentiable on some closed interval $[a,b]$ and has a derivative $f'$ which is not bounded. Is there a way to show that $f$ is non-Lipshitz?
If not, is there some intuitive counter example (like of the form $x^p\sin{\left(\frac{1}{x}\right)})$?
Yes, that conclusion holds. A proof by contraposition would be to show that (for a differentiable function $f$ on $[a, b]$)
$$ \text{$f$ is Lipschitz} \implies \text{$f'$ is bounded.} $$
So assume that $f$ is Lipschitz continuous with some Lipschitz constant $L$. Then $$ \left\vert \frac{f(x)-f(x_0)}{x-x_0} \right \vert \le L $$ for all $x, x_0 \in [a, b]$ with $x \ne x_0$. Taking the limit $x \to x_0$ it follows that $$ |f'(x_0)| \le L $$ for all $x_0 \in [a, b]$, i.e. $f'$ is bounded.