EDIT I feel it might be better to explain my motivation here. In fact, I want to prove the following result
$X$ is Banach, $M$ convex, $x\notin M$, then: $$d(x,M)=\sup_{f\in X^*,\|f\|=1}\{f(x)-\sup_{z\in M}f(z)\}.$$
And I did prove that $RHS\le LHS$, which was quite straightforward and didn't even require $X$ being Banach or $M$ being convex. The other direction however was much trickier and led to the below problem.
Let's assume $X$ is a normed space, $U$ convex and closed, and $V$ convex and compact. $U\cap V=\emptyset$. Clearly $d(U,V)>0$. Also, by Hahn Banach Theorem, $U,V$ can be strictly separated by a real-valued continuous functional $f$, that is, $$\inf_U f(x)-\sup_V f(x)>0.$$ Intuitively, there should be such a $f$ with $\|f\|=1$ that LHS actually equals $d(U,V)$, but I couldn't find any reference on this topic.
I then considered the deteriorated case where $V=\{0\}$. On one hand, by Hahn Banach I could find a functional $f$ with the unit norm such that $f(z)=d(0,U)$ where $z$ is the unique point at which $U$ is closest to the origin. The problem is $f$ may not separate $V$ from the origin. But on the other hand, for functionals separating the two, they may not attain $d(U,V)$. Is there a way to reconcile the two conditions to proceed, or any alternative method?