I am reading Functional Analysis by Conway. The location of that theorem is Chapter V, Prop 4.1. $\sigma(A^{**},A^{*})$ means weak star topology on $A^{**}$.
The prove is as follows:
Let $B$ be the $\sigma(A^{**},A^{*})$-closure of ball $A$ in $A^{**}$; clearly, $B \subseteq \text{ball}\, A^{**}${because ball $A^{**}$ is weak-* compact by banach Alaoglu's theorem}. If there is an $x_0^{**}$ in ball $A^{**} \text{\B}$, then the Hahn-Banach theorem implies there is an $x^*$ in $A$, and $\alpha$ in $\mathbb R$, and an $\epsilon >0$ such that $$\operatorname{Re}⟨x,x^*⟩<\alpha<\alpha+\epsilon< \operatorname{Re}⟨x^*,x_0^{**}⟩$$ for all $x$ in ball $A$.
And here I am unable to understand how to use Hahn-Banach theorem.
What I understood:-
First of all what is the meaning of $x^{**}$ belongs to B in the topology of $A^{**}$. Here $\sigma(A^{**},A^*)$ is the topology such that the collection of evaluation maps $${\{\hat{x^*}:A^{**} \rightarrow \mathbb{C}:\hat{x^*}(x^{**})=x^{**}(x^{*}) \}}$$ are continuous. Since $x_0^{**}$ not in B so there is no sequence $x_n$ in ball $A$ which converges to $x_0^{**}$ in $A^{**}$ with respect to $\sigma(A^{**},A^{*})$ topology i.e. for any sequence $x_n$ in $A$ there is some $x^*$ in $A^*$ such that $\hat{x^*}(\hat{x_n})$ does not converges to $x_0^{**}$. That I am able to understand but unable to understand how to get such $x^*$ as in book.
I cannot really follow your reasoning, but here is what's happening. The set $B$ is convex and closed in the weak$^*$-topology, and $x_0^{**}\not\in B$. Then we apply Hahn-Banach, to the sets $B$ and $\{x_0^{**}\}$, in the locally convex space $X=(A^{**},\sigma(A^{**},A^*))$. So there exists $\varphi\in X^*$ and $\alpha\in\mathbb R$, $\epsilon>0$, such that $\def\Re{\operatorname{Re}}$ $$ \Re\varphi(x)<\alpha<\alpha+\epsilon+\Re\varphi(x^{**}_0),\qquad\qquad x\in B. $$ If you look at Theorem V.1.3, you'll see that $X^*=A^*$. This means that there exists $x^*\in A^*$ such that $\varphi(x)=\langle x^*,x\rangle$.
As a comment, looking at the weak$^*$-topology the way you do is confusing (to me, at least). The weak$^*$-topology is just the topology of pointwise convergence: $\varphi_j\to \varphi$ means $\varphi_j(x)\to\varphi(x)$.