It is well known that, given a Banach space $Y$ and its dual $Y^*,$ a subset $G$ of $Y^*$ is $w^*-$compact iff is $w^*-$closed and norm- bounded.
Could you show an example of a normed space $Y$ and a $w^*-$compact set $G\subseteq Y^*$ such that $G$ is unbounded in the dual norm? Of course, $Y$ cannot be Banach.
Take $Y = \ell_0 (\Bbb{N})$ to be the space of all finitely supported sequences, equipped with the $\ell^\infty$-norm.
Take $\varphi_N : Y \to \Bbb{C}, (x_n)_n \mapsto N \cdot x_N$. It is easy to see $\varphi_N (x) \to 0$ for each fixed $x \in Y$ (in fact, $\varphi_N (x) = 0$ for $N \geq N(x)$ large enough).
But since $x_N = \delta_N \in Y$ with $\|x_N\|_{\ell^\infty} = 1$, we have $\|\varphi_N\| \geq |\varphi_N (x_N)| = N$, so that the set $G := \{\varphi_N : N \in \Bbb{N}\} \cup \{0\}$ is not bounded.
Finally, since $\varphi_N \to 0$ in the weak-$\ast$-topology induced by $Y$, it is not hard to see that $G$ is indeed weak-$\ast$-compact. In fact, if $(U_i)_i$ is an open covering of $G$, then $0 \in U_i$ for some $i$. Since $\varphi_N \to 0$, this implies $\varphi_N \in U_i$ for all but finitely many $N$. From here, it is not hard to extract a finite subcover.