I think I can show that for any function $f\in \mathcal{C}^2(\mathbb{R}^d)$ s.t. $f$ is Lipschitz with constant $L$, then $\nabla f$ is Lipschitz as well. This result relies on the fact that the first and second order gradients of $f$ exist and are continuous. I am not sure if indeed I show the desired result or if the result itself is true.
My proof attempt:
Suppose $f$ is as above. It is obvious $\nabla f$ is bounded by $L$, but suppose that $\nabla^2 f$ is unbounded. Unboundedness is equivalent to: $\forall M>0$, $\exists x$ s.t.\begin{equation}\|\nabla^2 f(x)\|>M.\end{equation}As $\nabla^2 f$ is also continuous, there is a sequence $\{x_k\}^\infty_{k=1}$ of lower bounded step-size (i.e. $\|x_k-x_{k+1}\|>\delta$) where $\forall M>0$ there exists $K$ s.t. $\forall k>K$ and $\forall t\in [0,1]$,\begin{equation}\|\nabla^2 f(tx_k + (1-t)x_{k+1})\|>M.\end{equation}(This is the part I am unsure about, but I believe it to be a consequence of the fact that all unbounded points are at $\infty$ in some sense). Now we choose $M=2L/\delta$ and apply the MVT to the points $x_k,x_{k+1}$ for $k>K$, to obtain\begin{equation}2L\geq \|\nabla f(x_k) - \nabla f(x_{k+1})\|= \|\nabla^2 f(x^\star)\| \|x_k-x_{k+1}\| > 2L.\end{equation}Where the RHS follows from the fact that $x^\star \in \{tx_k + (1-t)x_{k+1}: t\in[0,1]\}$. This produces a contradiction, from which follows that $\nabla^2 f$ is bounded and hence $\nabla f$ is Lipschitz.
Any help or clarification with this problem would be greatly appreciated.