Prove that a set which satisfies every nonempty subset contains a weak Cauchy sequence must be bounded

Question

I got stuck on this problem about weak convergence in normed space. This problem is exercise 9, page 263 in Functional Analysis of Erwin Kreyszig. So the question is basically on the title, let me state it clearly here:

Let $A$ be a set in a normed space $X$ such that every nonempty subset of $A$ contains a weak Cauchy sequence. Show that $A$ is bounded.

Here, a sequence $(x_n)$ is a weak Cauchy sequence if for every $f' \in X'$, the sequence $(f(x_n))$ is Cauchy in $\mathbb{R}$ or $\mathbb{C}$ ($X'$ is dual space of $X$: space of all bounded linear operators on $X$). I have no idea how to solve this problem, so I hope someone can give me a hint. Thanks so much.

Answer 1

If $x_n\in X$ and $||x_n||$ is unbounded then the Banach-Steinhaus Theorem, also known as the Uniform Boundedness Principle, shows that there exists $\Lambda\in X^*$ such that $\Lambda x_n$ is unbounded. So there is a subsequence $x_{n_j}$ with $|\Lambda(x_{n_j}|\to\infty$; that subsequence has no weakly Cauchy subsequence.