= Solution
Consider the restriction operator
$$
T:X\longrightarrow E^*,\qquad (Tx)(e)=e(x),\qquad a=x^{**}|_E.
$$
It is onto: otherwise its range, a <vector subspace> of the <finite-dimensional vector space> $E^*$, would have a nonzero annihilator in $E^{**}=E$. That annihilator would be an $e\in E$ vanishing on all of $X$, so $e=0$, a contradiction. Moreover $T$ is open. To see this directly, choose preimages of a basis of $E^*$; they define a linear right inverse $S:E^*\to X$, continuous because its domain is finite-dimensional. Small changes of an image can then be lifted by small changes using $S$.
Put $r=1+\varepsilon$ and $C=T(\{x:\|x\|<r\})$. It is open and convex in $E^*$. If $a\notin C$, the <Hahn-Banach separation theorem> gives a nonzero real <linear functional> on $E^*$, represented by some $e\in E$, such that
$$
a(e)\geq\sup_{z\in C}z(e)=r\|e\|.
$$
But $a(e)=x^{**}(e)\leq\|e\|$, contradicting $r>1$. Therefore $a\in C$, and the <finite-dimensional interpolation form of Goldstine's theorem> yields
$$
\boxed{\|x\|<1+\varepsilon,\qquad e(x)=x^{**}(e)\text{ for every }e\in E.}
$$
If $E=\{0\}$ one simply takes $x=0$.
To recover the <Goldstine theorem>, start with $x^{**}\in B_{X^{**}}$ and finitely many tests $e_1,\ldots,e_m\in X^*$. Apply the result to their span with a small parameter $\delta>0$. The resulting $x$ need not lie in $B_X$, but $v=x/(1+\delta)$ does. For every test,
$$
|e_j(v)-x^{**}(e_j)|=\frac{\delta}{1+\delta}|x^{**}(e_j)|\leq\delta\|e_j\|.
$$
Choosing $\delta$ sufficiently small puts $Jv$ in any prescribed basic weak-star neighborhood. This proves the asserted weak-star density of the closed <unit ball>. The reverse inclusion follows because $B_{X^{**}}$ is weak-star closed, being the intersection of the conditions $|x^{**}(e)|\leq\|e\|$ for $e\in X^*$.
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