Derivative has Linear Growth Implies Lipshitz

Question

Let $f\in C^\infty(\mathbb{R}^d)$. If $f$ has linear growth i.e

$$|\nabla f(x)|\leq C(|x|+1)$$

then is $f$ Lipshitz?

attempt at proof :

by Mean Value Theorem there exists $c\in (0,1)$ such that \begin{align*} |f(x)- f(y)| \leq |\nabla f((1-c)x+cy)||x-y|\leq & C(|x-y|+1)|x-y| \\ \leq& C(|x-y|^2+|x-y|) \\ \leq& C(2|x-y|^2+1). \end{align*}

Answer 1

Try $f(x) = x^2$ on $\mathbf R^1$. Then $|\nabla f(x)| = |f'(x)| = 2|x| \le 2(|x| + 1)$ for all $x$, yet $f$ is not Lipschitz.