When the Banach space $V^*$ is reflexive, we have the unit ball in $V^*$ is weak$^*$ sequentially compact.
For a Banach space $V^*$ that might not be reflexive, we have to assume that $V$ is separable, then the unit-ball in $V^*$ is weak$^*$ weak$^*$ sequentially compact result.
Question:
When the space is reflexive, what is the intuitive reason that we do not require the pairing space to be separable for the sequential compactness result?
Proof of 1 (followed from book):
We prove in terms of weak convergence instead of weak$^*$ convergence since the space is reflexive.
(a) First if $V = V^{**}$ is separable, then the result is true from (2).
(b) Now assume $V = V^{**}$ is not separable, given $\{v_n^*\}$ in the unit ball, if we let $W^*$ to be the clousure of the subspace generated by $\{v_n^*\}$, then $W^*$ is separable, reflexive and $W^{**}$ is also separable, reflexive (there is a ref in my book for this). Using the result of (a), we know there exists a weakly convergent subsequence under $\sigma(W^*, W^{**})$, since $V^{**} \subset W^{**}$ the subsequence is also convergent under $\sigma(V^*, V^{**}) =\sigma(V^*, V)$.
To answer my question, i guess that the result $W^{**}$ is separable might not hold when the space $V^*$ is not reflexive?
Thank you very much!