There is the following result for Banach spaces:
Let $X$ be a Banach space. Then a subset of the dual $X^*$ is bounded if and only if it is weakly* bounded.
To show that this result does not hold for normed spaces, consider the space $(c_{00} (\mathbb{N}), \| \cdot \|_1)$. For each $m \in \mathbb{N}$, define $$ f_m : c_{00} (\mathbb{N}) \to \mathbb{F}, \quad (a_n)_{n \in \mathbb{N}} \mapsto m \cdot a_m,$$ and let $A := \{ f_m \; | \; m \in \mathbb{N} \}$. I have shown that the set $A$ is weakly* bounded:
Since $c_{00} (\mathbb{N})$ forms a subspace of $\ell^1 (\mathbb{N}) \subset \ell^{\infty} (\mathbb{N})$, it holds, for all $(a_n)_{n \in \mathbb{N}}$, that $\sup_{n \in \mathbb{N}} |a_n| < \infty$, and hence that $$ \sup \{ | \langle (a_n)_{n \in \mathbb{N}}, f_m \rangle | \; | \; f_m \in A \} = \sup \{ | m \cdot a_m | \; | \; m \in \mathbb{N} \} < \infty $$ for all $(a_n)_{n \in \mathbb{N}}$. Thus $A$ is weakly* bounded.
However, I have difficulties showing that $A$ is not norm bounded. It is clear to me that if the uniform boundedness principle was applicable, then an application would show that $A$ was norm bounded.
Any help is appreciated.
The norm of $f_m$ is at least $m$ by taking the sequence $\delta_m$, which is only nonzero at $m$, where it has value $1$. Thus $A$ cannot be norm bounded, as it contains $f_m$ for all $m \in \mathbb{N}$.