Let $X$ be a normed space, $\bar{x} \in X,$ and $f: X\to \Bbb R$ be a convex functional that is continuous at $\bar{x}.$ It is well known that in this case, $\partial f(\bar{x})$ is nonempty and $w^*$- compact. In particular, if $X$ is Banach, it follows that $\partial f(\bar{x})$ is (dual norm) bounded as a consequence of the Uniform Boundedness Theorem. Hence the question:
Does there exists a normed space $X,$ a point $\bar{x}\in X,$ and a convex functional $f: X \to \Bbb R$ continuous at $\bar{x}$ such that $\partial f(\bar{x})$ is (dual norm) unbounded??
Of course, if it exists, $X$ cannot be Banach.
It is always bounded in this case. If you want an unbounded subgradient then you need a function that is not continuous at $\bar x$.
If $f$ is continuous at $\bar x$, there exists a $\delta>0$ such that $f(\bar x+y)<f(\bar x)+1$ for all $y$ with $\|y\| < \delta$.
Let $\xi\in\partial f(\bar x)$ be given. Then we have $$ \|\xi\| = \delta^{-1}\sup_{\|y\|<\delta} \xi(y) \leq \delta^{-1}\sup_{\|y\|<\delta} (f(\bar x+y)-f(\bar x)) \leq \delta^{-1} 1 = \delta^{-1}. $$ Therefore $\partial f(\bar x)$ is bounded.