If $\Omega \subset \mathbb{C} $ is open and connected and $ f:\Omega \to \mathbb{C}$ is an analytic function, such that $\,\,K=\overline{\text{Im}\, (\,f)} \subset \Omega$ is a compact set. Prove that $f$ has a unique fixed point.
First attempt. Let $K=\mathrm{Im}\,(\,f)$. If $f$ is not constant, then it is open and hence $$ \Omega\supsetneq K \supsetneq f(\Omega) \supsetneq f(K)\supsetneq\cdots\supsetneq f^{n}(\Omega)\supsetneq f^{n}(K)\supsetneq\cdots. $$ Then if $K_n=f^{n}(K)$, clearly $L=\bigcap f^{n}(\Omega)=\bigcap K_n\ne\varnothing$. It suffices to show that $L$ is a singleton.
Consider the well-defined sequence: $$ f_1=f,\,\,f_{n+1}=f\circ f_{n}, \quad n\in\mathbb N. $$ Clearly, $\{f_n\}$ is a normal family, as they are all bounded, and hence there exists a locally uniformly converging subsequence $f_{n_k}\to g$, where $g: \Omega\to\mathbb C$ holomorphic. As observed in the OP $$ \Omega\supsetneq K \supsetneq f(\Omega) \supsetneq f(K)\supsetneq\cdots\supsetneq f_{n}(\Omega)\supsetneq f_{n}(K)\supsetneq\cdots. $$ Hence, for a limit $g$ of a locally uniformly converging subsequence $\{f_{k_n}\}$, we should have that $$ g(\Omega)\subset L=\bigcap_{n=1}^\infty f_n(\Omega)=\bigcap_{n=1}^\infty f_n(K). $$ Clearly, $L$ is compact, as an intersection of compact sets. We shall now show that $L\subset g(\Omega)$, and hence, as $\Omega$ is connected, due to the Open Mapping Theorem, $g$ is constant and $L$ is a singleton.
Let $w\in L$. Then, for every $n\in\mathbb N$, there exists a $z_n\in\Omega$, such that $w=f_n(z_n)$. But $f_{n+1}(z_{n+1})=f_{n}\big(f(z_{n+1})\big)$, and hence the sequence $\{z_n\}$ could be chosen so that $\{z_n\}_{n=m+1}^\infty \subset f_m(K)$. Now clearly, $\{z_{k_n}\}$ possesses a converging subsequence $\{z_{\ell_n}\}$, with $z_{\ell_n}\to\zeta_0\in K\subset\Omega$. We have that $$ w=f_{\ell_n}(z_{\ell_n})=\lim_{n\to\infty}f_{\ell_n}(z_{\ell_n})=g(\zeta_0), $$ and hence $w\in g(\Omega)$.
We have shown that $L=\{z_0\}$. Clearly, $\,f(L)=L$, and if $\hat z_0$ were another fixed point, then $\hat z_0 \in L$.