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A norm-closed finite-codimensional vector subspace $Y\subseteq X^{\ast }$ that is weak-star dense is a norming subspace. Its annihilator $F=Y^{\perp }\subseteq X^{\ast \ast }$ is finite-dimensional and disjoint from $JX$. The positive distance $d$ of the unit sphere from $F$ gives a triple-dual linear functional vanishing on $F$ and taking a value at least $d$ at a unit vector. Apply the finite-dimensional interpolation form of Goldstine's theorem to the Banach space $X^{\ast }$, then normalize and take a supremum to obtain norming constant $d$.

Consider the restriction operatorIt is onto: otherwise its range, a vector subspace of the finite-dimensional vector space $E^{\ast }$, would have a nonzero annihilator in $E^{\ast \ast }=E$. 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 $E^{\ast }$; they define a linear right inverse $S:E^{\ast }\to X$, continuous because its domain is finite-dimensional. Small changes of an image can then be lifted by small changes using $S$.

Put $r=1+\epsilon$ and $C=T(\{x:\Vert x\Vert <r\})$. It is open and convex in $E^{\ast }$. If $a\notin C$, the Hahn-Banach separation theorem gives a nonzero real linear functional on $E^{\ast }$, represented by some $e\in E$, such thatBut $a(e)=x^{\ast \ast }(e)\le \Vert e\Vert$, contradicting $r>1$. Therefore $a\in C$, and the finite-dimensional interpolation form of Goldstine's theorem yieldsIf $E=\{0\}$ one simply takes $x=0$.

To recover the Goldstine theorem, start with $x^{\ast \ast }\in B_{X^{\ast \ast }}$ and finitely many tests $e_1,\dots ,e_m\in X^{\ast }$. 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,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 $B_{X^{\ast \ast }}$ is weak-star closed, being the intersection of the conditions $|x^{\ast \ast }(e)|\le \Vert e\Vert$ for $e\in X^{\ast }$.