Does this proof of the sequential completeness of a reflexive space require the Eberlein–Šmulian theorem?

Question

I'm self-teaching functional analysis through Conway's book at the moment, and can't seem to get one of the steps in this proof:

The statement that $\{ x\in X : ||x||<M\}$ implies the sequence $\{x_n\}$ has a cluster point is throwing me off. I can't seem to find a way to justify this without invoking the Eberlein–Šmulian theorem - which Conway doesn't mention in his book until a few chapters later. My question is - am I missing something? The definition Conway gives for a Cauchy sequence in this situation is given below:

Is there a way to see thus statement without using such heavy machinery? Or is this just a small oversight in his book?

Edit: the problem is that weak compactness implies weak sequential compactness here. I should’ve specified that fact that ballX is weakly compact isn’t the problem

Answer 1

Weak sequential compactness isn’t actually required here. K a compact space $\implies$ every net in K has a cluster point.

Answer 2

I haven't checked Conway to see what has been proven up until this point, but this assertion can be proved using the following elementary argument:

Let $\mathscr Y = \overline{\operatorname{span}}\{x_n\}$ be the closed span of $\{x_n\}$ in $\mathscr{X},$ which is a separable reflexive Banach space with the inherited norm. Then I claim $\{ y \in \mathscr{Y} : \lVert y \rVert \leq M\}$ is weakly sequentially compact; this can either be proven directly by means of a diagonal argument, or by noting the set is weakly metrisable. Hence we infer there is $x \in \mathscr{Y} \subset \mathscr{X}$ such that $x_n \rightharpoonup x$ weakly in $\mathscr{Y},$ and it is easy to see this also implies weak convergence in $\mathscr{X}.$