Suppose $\nabla^2 V$ has a negative eigenvalue $-\lambda_0$ at point $x_0$, with eigenvector $v_0$, consider the modified flow

$$ Y' = - \nabla V(Y) + \nabla V(x_0) $$

which has $x_0$ as a stationary point. This new flow is a constant shift of the original flow and has the same Lipschitz properties.

In a neighborhood of $x_0$, consider the modified flow with initial data $x_0 + \epsilon v_0$ for some sufficiently small $\epsilon\ll \lambda_0$. The flow looks like

$$ (Y - x_0)' = -\nabla^2 V(x_0) \cdot (Y - x_0) + O(|Y-x_0|^2) $$

and so for short times the solution is well-approximated by

$$ x_0 + \epsilon v_0 e^{\lambda_0 t} + O (\epsilon^2) $$

and so violates the 1-Lipschitz assumption.

The case where $\nabla^2 V \geq \lambda \mathrm{Id}$ is similar.

Doesn't this follow from dependence on initial data?

Consider the flow mapping $\Phi(t,X)$ which solves

$$ \frac{d}{dt}\Phi(t,X) = - \nabla V(\Phi(t,X)) $$

so taking the derivative in $X$ we have

$$ \frac{d}{dt} \partial_X \Phi(t,X) = - \nabla^2 V(\Phi(t,X)) \cdot \partial_X \Phi(t,X) \\= - \nabla^2 V(X) \cdot \partial_X \Phi(t,X) + O(t) \cdot \partial_X \Phi(t,X)$$

So if $-\nabla^2 V(X_0)$ has negative eigenvalue $-\lambda_0$ with eigenvector $v_0$, taking the partial in the $v_0$ direction gives

$$ \partial_{v_0} \Phi(t,X_0) = e^{\lambda_0 t} v_0 + O(t^2) $$

For $t>0$ sufficiently small you guarantee that $$ |\partial_{v_0} \Phi(t,X_0) | \geq (1 + \frac{\lambda_0}{2}t) |v_0| $$ showing that the solution map cannot be 1 Lipschitz.