In order to be as clear as possible, I've taken the liberty of TeXing (Tikzing?) up the sort of image in question.
Here, $\gamma$ is some path in the complex plane, the disk $\Delta(z_0,2r)\subset{U}$ where $U=\mathbb{C}\sim|\gamma|$ and $|\gamma|\subset\Delta(0,s)$.
I ask this because a proof in my book says that, if $z$ lies in the punctured disk $\Delta^*=\Delta^*(z_0,r)$ (i.e., the set given open disk of radius $r$ less $\{z_0\}$) and $\zeta$ on $|\gamma|$, then $|\zeta-z_0|\ge{r}$ and $|\zeta-z|\ge{r}$.
What puzzles me is the use of the $\ge$ symbols, as it seems to imply that the set defined as the distance between the boundary of $\Delta(z_0,2r)$ and the set of points in $\Delta^*$ is closed, as it contains all of its limit points.
I cannot justify this, as $\gamma$ is permitted to be within $2r$ of $z_0$, but $z$ cannot be within $r$ of the boundary of the disk $\Delta(z_0,2r)$. Certainly any point located a distance $r$ away from $z_0$ is an accumulation point of the punctured disk, but not an accumulation point contained within it. Correct?
Or is it enough to say that, given some sequence $\langle{z_n}\rangle$ such that $|z_n-z_0|\rightarrow{r}$ as $n\rightarrow\infty$, the value $r$ must be included in the set of distances between $\Delta^*$ and $2r$?
Compactness helps simplify this. Distance is a continuous function of two variables. So, $d(\xi,\eta)$ is continuous as $\xi$ varies over $\gamma$ and $\eta$ varies independently over $\Delta(z_0,r)$. But the Cartesian product $\gamma \times \Delta(z_0,r)$ is compact, because $\gamma$ is compact and $\Delta(z_0,r)$ is compact. The continuous image of a compact set is compact. So the set of values of $d(\xi,\eta)$ is compact, and therefore is a closed and bounded subset of $\mathbb{R}$.