Assume that $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a convex function and $\mathrm{grad}f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is its gradient. For $\varepsilon>0$, is the set $$ \{ x\in\mathbb{R}^n \, \mid \, \Vert \mathrm{grad} f(x)\Vert < \varepsilon\}$$ convex?
When $n=1$, this is true since $\mathrm{grad} f(x)=f'(x)$ is an increasing function which implies that the set in question must be an interval.
The assertion is not true. Consider the convex function $$ f(x,y) := \frac14 (x^4 + y^4) - x - y.$$ Then $$ \nabla f(x,y) = \begin{pmatrix} x^3 -1 \\ y^3 - 1\end{pmatrix}.$$ Consequently, $$ \|\nabla f(1,0)\| = \|\nabla f(0,1)\| = 1,$$ but for the midpoint we have $$ \|\nabla f(1/2, 1/2)\| \approx 1.23,$$ which is bigger.
Here are some contour lines of $\|\nabla f(\cdot)\|$: