In the below attached picture, there is, solution provided for showing
$f(z) =\frac{1}{z}$ is not uniformly continuous un the region $|z|<1$
By definition of uniform continuity, we have for given any $\epsilon >0$ there exists $\delta >0$ (depending on $\epsilon$ only not on particular point $z_0$) such that, $|f(z) -f(z_0) |< \epsilon$ whenever $|z-z_0|<\delta$.
So clearly definition does not say anything about $\delta$. So how can they choose $\delta$ between $0$ & $1$. Why can't be $\delta>1$?
Since $|z-z_0|$ denotes distance between point $z$ & $z_0$ and as these two points lies in the region $|z|<1$ i.e. set of all points lying inside the circle of radius $1$ with center at origin. hence distance between $z$ and $z_0$ can be greater than $1$. So how can they choose $\delta < 1$?
The argument is as I said. If $\vert z-z_0\vert<\delta_\epsilon$ implies $\vert f(z)-f(z_0)\vert<\epsilon$, while $\delta_\epsilon\geq1$, then $\vert z-z_0\vert<\eta_\epsilon$ implies $\vert f(z)-f(z_0)\vert<\epsilon$ for any $0<\eta_\epsilon<1$. Now just work with this $\eta_\epsilon$ instead. They can just always replace $\delta_\epsilon$ with an $0<\eta_\epsilon<1$.
Later Edit:
There is indeed a problem at sets whose closure contain $z=0$, for the reason you said. One cannot extend $f$ to $0$ continuously, but $f$ is continuous at any $A\subseteq \mathcal{C}\setminus \{0\}$. Moreover, $f$ is uniformly continuous on such an $A$, if $A$ is compact. Therefore, any counterexamples to uniform continuity will have to come from points close to $z=0$. I hope this what you were asking about in the comment.