I stumbled upon the following problem, which I am yet unable to solve.
Let X be a normed space and let $C \subset X^*$ be a nonempty, $w^∗$ (weak-star) closed set. Show that $C$ is proximinal, that is, for any given $x^* \in X^*$, there exists $u \in C$ such that $ ||x^* - u || = d(x^* , C)$.
Any hints?
I think I've got it. Let $x^* \in X^*$ and $d=d(x^*,C): = \inf \{ ||x^*- z|| \colon z \in C\}$. By the definition of infimum, for any $ n \in \mathbb N$, we have that $ B_{d+1/n}(x^*)\cap C:= \{ z \in C \colon || z-x^*|| \leq d+1/n \} \neq \emptyset$. Let $$ A_n = B_{d+1/n} (x^*)\cap C , \ n \in \mathbb N.$$ The family $(A_n)_n $ consists of $w^*$-closed sets and it can easily be checked that it has the finite intersection property. Since $\ B_{d+1}(x^*) \supset A_1 \supset A_2 \supset \dots \ $ and $B_{d+1}(x^*)$ is $w^*$ compact (Alaoglou's theorem), we must have that $ \bigcap_{n \in \mathbb N} B_{d+1/n}(x^*) \cap C \neq \emptyset $. It's not hard to see that for any $z \in \bigcap_{n \in \mathbb N} B_{d+1/n}(x^*) \cap C ,\ $ $||x^*-z||=d$.