How to show that a weak-star convergent sequence is norm bounded?

Question

In this MSE post, the author stated that

as a consequence of uniform boundedness principle, a weak-star convergent sequence is norm bounded.

I tried to prove it but to no avail. Any hint is appreciated.

Answer 1

$\textbf{Claim}$: Let X be a banach space and $K^* \subset X^*$ a weak-star compact set. Then $K^*$ is norm bounded.

$\textbf{Proof}$: Let $$e \colon X \to X^{**}, \ \ x \mapsto e(x) \in X^*, \text { with } \langle e(x),x^* \rangle = \langle x^*,x \rangle, \ \text{ for any } x^* \in X^* $$ be the canonical isometric embedding.

For every $ x \in X $, the functional $ e(x) \colon X^* \to \mathbb R $ is (by definition) $ w^*-$continuous and therefore the set $ e(x) (K^*) \subset \mathbb R $ is compact. Hence, $ \sup_{x^* \in K^*} | \langle e_x ,x^* \rangle | < \infty $ and so, $ \sup_{x^* \in K^*} | \langle x^*,x \rangle|< \infty$. In other words, the family of operators $ \{ x^* \}_{x^* \in K^*} $ is pointwise bounded (for every $ x \in X $) and since $ X $ is Banach, applying the uniform boundedness principle, we have that $ \sup_{x^* \in K^*} ||x^*|| < \infty$.

Now, if $(x^*_n)_n \subset X^*$ is a weak-star convergent sequence with limit $x^*$, then the set $\{ x^*_n,x^*\}_n$ is weak-star compact.