A Banach space is reflexive if and only if its dual is reflexive How to show that a Banach space $X$ is reflexive if and only if its dual $X'$ is reflexive?

Here's a different, more geometric approach that comes from Folland's book, exercise 5.24
Let $\widehat X$, $\widehat{X^*}$ be the natural images of $X$ and $X^*$ in $X^{**}$ and $X^{***}$.
Define $\widehat X^0 = \{F\in X^{***}: F(\widehat x) = 0  \text{ for all } \widehat x \in \widehat X\}$
1) It isn't hard to show that $\widehat{X^*} \bigcap \widehat X^0 = \{0\}$.
2) Furthermore, $\widehat{X^*} + \widehat X^0 = X^{***}$. To show this, let $f\in X^{***}$, and define $l \in X^*$ by $l(x) = f(\widehat x)$ for all $x\in X$.
Then $f(\phi) = \widehat l(\phi) + [f(\phi) - \widehat l(\phi)]$. 
Clearly $\widehat l \in \widehat{X^*}$, and we claim $f - \widehat l \in \widehat X^0$. Let $\widehat x \in \widehat X$. Then $f(\widehat x) - \widehat l ( \widehat x) = f(\widehat x) - \widehat x (l) = f(\widehat x) - l(x) = 0$
Now that 1) and 2) are verified, we prove the claim:
If $X$ is reflexive, then $\widehat X^0 = \{0\}$, and so $X^{***} = \widehat{X^*}$, so $X^*$ is reflexive.
If $X^*$ is reflexive, then $X^{***} = \widehat{X^*}$, so $\widehat X^0 = \{0\}$. Since $\widehat X$ is a closed subspace of $X^{**}$ (on assumption $X$ is Banach), if $\widehat X$ were a proper subspace of $X^{**}$, we would be able to use Hahn-Banach to construct an $F \in X^{***}$ such that $F$ is zero on $\widehat X$ and has ||F|| = 1. This, however, would contradict $\widehat X^0 = \{0\}$. So we conclude $\widehat X  = X^{**}$.
