Is there an analytic function that satisfies this equality? Let $U$ be an open subset of $\mathbb{C}$ that is bounded. Let $z_1,z_2 \in U$ be distinct. Is there an analytic function $f: U \rightarrow \mathbb{D}$ satisfying $|f(z_1)-f(z_2)| = \sup \{|g(z_1)- g(z_2)| : g: U \rightarrow \mathbb{D} \text{ analytic} \}$?
 A: By Montel's theorem, the set $A_c(U, \mathbb D)$ of analytic functions $h:U \to \mathbb {\bar D}$ is a compact metric space and it contains non-constant functions since $U$ bounded
(exhaust $U$ with countable increasing bounded open subsets with compact closure in $U$ so $\bar U_n \subset U,  U_1 \subset U_2 ..\subset U_n \subset ..., \cup U_n=U$ and take the metric $d(f,g)=\sum 2^{-n}\frac {||f-g||_n}{1+{||f-g||_n}}$, where $||f|_n=\max_{\bar U_n}|f|$)
Fix $z_1,z_2$ and consider $L:A_c(U, \mathbb D) \to [0, 2], L(f)=|f(z_1)-f(z_2)|$; $L$ is continuos hence it attains the maximum at some $f \in A_c$; if $U$ is connected we are done since $f$ cannot be constant as $z_1 \ne z_2$, so $f(U) \subset \mathbb D$ by maximum modulus. 
On the other hand if $U$ is not connected and $z_{1,2}$ are in different connected components of $U$ we definitely can have the maximum attained at functions that are say $\pm 1$ on those components and since for any function $f: U \to \mathbb D$, $L(f) < 2$ we do not have an $f$ as required, but that imho is more of a degenerate case, so in the generic $U$ connected case we are good and we have $f$.
