Let $\Omega\subset\mathbb{R}^n$ be an open convex set. Let $f:\Omega\to\mathbb{R}$ be a convex function. We know that $f$ is differentiable almost everywhere. The subdifferential of $f$ at $x\in\Omega$ is defined as $$ \partial f(x)=\{g\in\mathbb{R}^n:f(y)\ge f(x)+g^T(y-x)\ \forall\,y\in\Omega\}. $$
My question: Is it true that for every $x\in\Omega$, for every boundary point $g$ of $\partial f(x)$, there always exists a sequence $\{x_n\}\subset\Omega$ such that $f$ is differentiable at each $x_n$, $x_n\to x$, and $\nabla f(x_n)\to g$?
My attempt:
This is not true for general $g\in\partial f(x)$. For example, in $\mathbb{R}$, the convex function $f(x)=|x|$ is differentiable on $\mathbb{R}\backslash{\{0\}}$ and $\partial f(0)=[-1,1]$. Since $f'(y)\in\{-1,1\}$ for all $y\neq 0$, then there does not exist $\{x_n\}\subset\mathbb{R}\backslash\{0\}$ such that $x_n\to 0$ and $\nabla f(x_n)\to 0\in\partial f(0)$.
I believe the statement is true at least in the speical case of $\mathbb{R}$. Since $\partial f(x)\subset\mathbb{R}$ is closed and convex, then it must be an interval. So I suppose it should be true that $$ \partial f(x)=\left[\liminf_{y\to x}\frac{f(y)-f(x)}{y-x},\limsup_{y\to x}\frac{f(y)-f(x)}{y-x}\right]. $$ This should help prove the statement.
I am a beginner in convex analysis. Any help would be appreciated.