Let $X$ be a normed space and $C$ an open convex subset of $X$. Let $f: C \to \mathbb R$ be convex. If $f$ is continuous at some point in $C$, then $f$ has non-empty subdifferential at all points in $C$. If $X$ is finite-dimensional, then $f$ is continuous.
Could you provide an example of a convex function $f$ that has empty subdifferential at some point?
Here is a counterexample for which $f$ has some "problems", since it is defined on an incomplete space.
We take $X = \ell^1$, but we equip it with the norm from $\ell^2$, i.e., $$ \|x\|_X := \left( \sum_{n=1}^\infty x_n^2 \right)^{1/2}.$$ Now, we define $f(x) := \| x \|_{\ell^1} = \sum_{n=1}^\infty |x_n| \in \mathbb R$ for all $x \in X$. Then, $f \colon X \to \mathbb R$ has the following properties: convex (clear), lower semicontinuous (using, e.g., Fatou's lemma) but discontinuous everywhere (rather straightforward). Moreover, one can show (recall that $X^*$ can be identified with $\ell^2$) $$ \partial f(x) = \{ y \in \ell^2 \mid y_n \in \partial |\cdot|(x_n) \;\forall n \in \mathbb N\} $$ and this, in particular, yields $$ \partial f(x) \ne \emptyset $$ if and only if $x \in c_c$ (subspace of finite sequences).
In the special case when $f:\mathbb{R}^d \to \mathbb{R}$ is convex, the set-valued map $\partial_Pf: \mathbb{R}^d \to \mathcal{B}(\mathbb{R}^d), x \mapsto \partial_Pf(x) $ takes nonempty, compact, and convex values and is upper semicontinuous and locally bounded.
From Discontinuous Dynamical Systems, Proposition 9, Point 2.
I formalize @daw's comment as below.
Let $f:X \to \mathbb R$ be a discontinuous linear functional. We will prove that $\partial f (a) = \emptyset$ for all $a \in X$. Assume the contrary that $x^* \in \partial f (a)$ for some $a \in X$. Then $$ f(x)-f(a) =\langle f, x-a \rangle \ge \langle x^*, x-a \rangle \quad \forall x \in X. $$
It follows that $x^* \le f$. We define $g:X \to \mathbb R$ by $g(x) :=f(x)-f(0)$. Then $g$ is linear. By this result, $g=x^*$. Hence $g$ and thus $f$ are continuous. This is a contradiction.