In the proof of the Eberlein-Smulian theorem

Question

I wanna show the following version of the Eberlein-Smulian theorem in the book <> by Joseph Diestel.

Theorem)$X$ is a Banach space, $B \subset X$ is a bounded subset. Then the following statements about B are equivalent: 2. $B$ contains a countable subset $C$ with no weak limit point in $X$. 4. $B$ is not weakly sequentially compact in $X$. (I omit 1 and 3 because these are not related to my question)

I wanna prove 4 to 2:

$B$ is not weakly compact, there is a sequence $(y_n)$ in $B$ such that $y_n$ has no weakly convergent subsequence in X, hence no norm convergence subsequence. Passing to subsequence, may assume $(y_n)$ is norm discrete.

After that, the author assume 2 is false and continue the proof, but I think once we choose such $(y_n)$, then $\{y_n\}$ is a countable subset of $B$ not containing weak limit points. Where did I wrong?

Answer 1

Just knowing that $(y_n)$ is norm-discrete does not allow us to conclude that the set $\{y_n\}$ has no weak limit points. For example, the standard orthonormal basis $(e_n)$ of $\ell^2$ is a norm-discrete sequence, but the set $\{e_n\}$ has a weak limit point, namely $0$.

(Of course, this sequence $(e_n)$ is weakly convergent, so it's not a "counterexample to the theorem" — my point is, norm-discreteness does not immediately bring us to the desired conclusion.)