Claim: If $X$ is a normed space, and identify $X$ as a subspace of $X^{**}$. Then $ball(X)$ is $\sigma(X^{**}, X^*)$ - dense in $ball(X^{**})$. Here, $ballX$ is a closed unit ball in $X$, and $ballX^{**}$ similar.
A part of the proof (Conway): Let B be the closure of $ballX$ in $\sigma(X^{**}, X^*)$. Suppose there exist some $x_0^{**} \in ball(X^{**}) - B $. Then, by the Hahn-Banach theorem, there exist an $x^* \in X^*, \alpha \in \mathbb{R}, \epsilon >0$ such that $$Re \, x^*(x) < \alpha < \alpha + \epsilon < Re \, x_0^{**}(x^*)$$ for all $x \in ballX$.
My question: Exactly how this implied by the Hahn Banach?
My attempt: By the Hahn-Banach (exactly, by separation theorem on locally convex space), there exist $\lambda \in (X^{**})^*$ such that $$ Re \, \lambda(x) < \alpha < \alpha + \epsilon < Re \, \lambda(x_0^{**}) $$ for all $x \in B$. If we redefine $x^* : = \lambda |_X \in X^{*}$ to obtain desired $x^*, $ then since $ball X \subseteq B,$ so we obtain
$$Re \, x^*(x) < \alpha < \alpha + \epsilon < Re \, \lambda(x_0^{**}) $$ for all $x \in ball X$. But I really don't understand why $x_0^{**}(x^*) = \lambda(x_0^{**})$, or equivalently by definition of our $x^*,$ $x_0^{**}(\lambda|_X) = \lambda(x_0^{**})$.
Thank you!