Let $JT$ denote the James Tree space, $Q: JT \longrightarrow JT\,^{\prime\prime}$ the canonical embedding, $D_\ast$ a countable norm-dense subset of $JT$, let $D= Q(D_\ast)$ and let $E = JT\,^\prime$. Note that $E$ is nonseparable in the norm topology and that the $w^\ast$-sequential closure of $D$ in $E^{\prime}$ is $E^{\prime}$ since $Q(JT)$ is $w^\ast$-sequentially dense in $JT\,^{\prime\prime}$ (for the last claim, see in particular Corollary 2 of Lindenstrauss and Stegall Examples of separable spaces which do not contain $\ell_1$ and whose duals are non-separable, Studia Math. 54 (1975), p.81--105).
I should mention that until the appearance of the Lindenstrauss-Stegall result cited above, it seems to have been open since the time of Banach whether there could exist a separable Banach space that has nonseparable dual and is $w^\ast$-sequentially dense in its bidual.
Eberlein -Smulian is usually stated in the form:
Let $A$ be a subset of a Banach space $X$. Then, $A$ is weakly compact
if and only if $A$ is weakly sequentially compact.
The form you gave is a consequence of this, as the unit ball is weakly compact if and only if the space is reflexive. But, since the form you gave says absolutely nothing about non-reflexive spaces, it is actually weaker than E-S.
Q1) The answer is true in general, by the (stronger) ES, and it does not require reflexivity.
Note that, if I remember right, the key for the proof is the following:
If $E$ is a separable Banach space, then $E*$ contains a countable total set. In this case, the weak topology becomes metrisable on weakly compact sets.
Now if $x_n$ is any sequence, the subspace spanned by $\{ x_n \}$ is separable, and the trick above shows that weak-compactness implies weak sequential compactness.
The other implication is a bit trickier, it usually is done by showing that if $A$ is weakly sequentially compact it is bounded and the weak* closure of $A$ is included in $E$.
Best Answer
No, not necessarily. Any reflexive Banach space has a weakly compact unit ball; so, by the Eberlein-Šmulian Theorem, any reflexive Banach space has a weak* sequentially compact unit ball.
Also, and more generally, it follows from Rosenthal's $\ell_1$ Theorem (a Banach space $X$ does not contain $\ell_1$ isomorphically if and only if every every bounded sequence in $X$ has a weakly Cauchy subsequence) that if $X^*$ does not contain $\ell_1$ isomorphically, then the unit ball of $X^*$ is weak* sequentially compact.