$W^{\perp}$ is weak* closed I was looking at this exercise but I'm confused

Let $V$ be a Banach space and let $W$ be a subspace of $V.$ Show that $W^{\perp}\subset V^*$ ($W^{\perp}$ annihilator of $W$) is weak* closed .

Sice $W^{\perp}$ is closed if $f_n\rightarrow f,$ then $f\in W^{\perp}$ but $f_n\rightarrow f$ $\Rightarrow$ $f \rightharpoonup f_n$ $\Rightarrow$ $f_{n} \stackrel{*}{\rightharpoonup} f$
but can I deduce that $W$ is weak* closed?
 A: It is possible to proof that every annihilator of a subspace is w*-closed. We shall note $\widehat{W}$ its identification seen in the bidual, $\widehat{W}\subset X^{**}$ (using the pointwise identification $\hat{x}: X^* \to \mathbb{K}$ defined by $\hat{x}(f):=f(x)$, for all $f \in X^*$). Now, we have the following identities
$\displaystyle W^{\perp} = \{{x^*}\in X^*: \ x^*(x)=0, \ \forall x \in W\} = \{ x^* \in X^*: \ \hat{x}(x^*)=0, \ \forall\hat{x}\in \widehat{W} \} = \bigcap_{\hat{x}\in\widehat{W}} \ker(\hat{x})$.
The last identity shows that the annihilator is an intersection of $w^*$-closed subset, since each $\hat{x}$ is $w^*$-continuous (the kernel of a $w^*$-continuos function is clearly $w^*$-closed).
A: Hint:
If $x^*\notin W^\perp$ then, there is $u\in W$ such that $|x^*(u)|>0$.
As the map $\hat{u}:y^*\mapsto y^*(u)$ is weak* continuous (by definition of the weak* topology),  the set $U:=\{y^*\in X^*: |\hat{u}(y^*)|>0\}=\hat{u}^{-1}(\mathbb{C}\setminus\{0\})$ is
weak* open. This means that $U\subset X^* \setminus W^\perp$.
