Every nonzero element in a Banach space has a norming extreme point For any Banach space $F,$ let $B_F$ be the closed unit ball of $F,$ that is, $B_F = \{x\in F: \|x\| \leq 1\}.$ Also, let $ext B_F$ be the set of extreme points of $B_F$ (Recall that $x$ is an extreme point of $B_F$ if $x = \frac{1}{2}(x_1+x_2)$ for some $x_1,x_2\in B_F$ implies that $x= x_1=x_2.)$
Let $F^*$ be the continuous dual space of $F.$

Question: Let $F$ be a Banach space and $x\in F\setminus\{0\}.$ Is it true that there exists $x^*\in ext B_{F^*}$ such that $x^*(x) = \|x\|?$

I think it is true. Assume that $\|x\| = 1.$
Consider 
$$S = \{x^*\in B_{F^*}: x^*(x) = 1\}.$$
By Hahn-Banach Theorem, $S$ is nonempty. 
Clearly $S$ is convex. 
I would like to show that $S$ is weak-star closed in $B_{F^*}$ so that it is weak-star compact (Banach-Alaoglu states that $B_{F^*}$ is weak-star compact).
Then by Krein-Milman Theorem, $S$ has an extreme point, say $z^*.$
It is easy to see that $z^*$ is also an extreme point of $B_{F^*}.$
Such $z^*$ is our desired bounded linear functional.
However, I am not sure how to show that $S$ is weak-star closed.
Any hint is appreciated.


I would like to show that $S$ is weak-star closed in $B_{F^*}$ so that it is weak-star compact 

For a fixed $x\in F$, the map
$$
 \phi_x : F^*\to\mathbb K,\;
 x^*\mapsto x^*(x)
$$
is continuous from the weak-star topology of $F^*$ to the ordinary topology of $\mathbb K$, where $\mathbb K\in\{\mathbb R,\mathbb C\}$.
This is because the weak-star topology of $F^*$ is the coarsest topology such that all functionals $\phi_y:F^*\to\mathbb K$ 
for $y\in F$ are continuous
(this is sometimes even used as the definition of the weak-star topology).
Therefore, the preimage $\phi_x^{-1}(\{1\})$ of the closed set $\{1\}$
is weak-star closed.
It follows that the set $S= B_{F^*}\cap \phi_x^{-1}(\{1\})$ is weak-star closed.
