It is well-known that differentiability implies the continuity. I am looking for the converse and found the following result.
Theorem: (Prop 2.36 page 85 [1]) If the convex function $f : X \rightarrow (-\infty, +\infty]$ is continuous at $x_0$, then $f$ is subdifferentiable at this point.
Here are some of their main arguments.
Proof. Let $H$ be the convex epigraph of the function $f$. Since $f$ is continuous and "bla-bla-bla", there exists a closed supporting hyperplane of $H$ which passes through $(x_0,f (x_0))$, i.e. there exists $(x_0^*, \alpha_0)\in X^*\times \mathbb{R}$ such that $$\alpha_0[f(x)-f(x_0)]+ \langle x_0^*, x-x_0\rangle \geq 0, \forall x\in \text{dom} f$$ that is $$f(x)\geq f(x_0)+ \langle \frac{-x_0^*}{\alpha_0}, x-x_0\rangle, \forall x\in \text{dom} f$$ So $-x_0^*/\alpha_0$ is a sub-gradient, thus, $f$ is subdifferentiable at $x_0$.
My question is how we know that $$\alpha_0> 0?$$
Assume $\alpha=0$. Then $$ \langle x_0^*, x-x_0\rangle \ge 0 $$ for all $x\in \ dom \ f$. The point $x_0$ is an interior point of $dom \ f$, since $f$ is assumed to be continuous at $x_0$. Hence $x-x_0$ can be taken from a small ball around zero. But then $x_0^*$ has to be zero. This is a contradiction to $(\alpha,x^*) \ne (0,0)$.
I suspect that the original inequality is $$ \alpha (r - f(x_0)) + \langle x_0^*, x-x_0\rangle \ge 0 $$ for all $(x,r) \in epi \ f$. And not the one in the question. Setting $x:=x_0$ and $r:=f(x_0)+1$ implies $\alpha \ge0$.