I've been looking for a solution of problem:
Let $U\subset\mathbb{C} $ be a open set and $\Gamma:[0,1]\to U$ is a continuous closed path (i.e. starts and ends at the same point. Then $\Gamma$ is homotopic to a polygonal path.
By definition, two closed continuous paths $\Gamma_0,\Gamma_1$ are homotopic in $U$, if there exists a continuous function $\gamma:[0,1]\times[0,1]\to U$ such that $\gamma(0,t)=\Gamma_0(t)$, $\gamma(1,t)=\Gamma_1(t)$ and $\gamma(s,0)=\gamma(s,1)$ for every $s\in(0,1)$.
I already found some answers, but no reliable proof that works also for ugly continuous paths. I tried to do it myself:
Denote $\Gamma^{\epsilon}:=\{z\in\mathbb{C};\;dist(z,\Gamma([0,1]))<\epsilon\} $. Since $\Gamma([0,1])$ is compact and $U$ open, we see that $\Gamma^{\epsilon}\subset U$ for some $\epsilon>0$. For this $\epsilon$ by uniform continuity of $\Gamma$, there exists $\delta>0$ such that if $|x-y|<\delta$ then $|\Gamma(x)-\Gamma(y)|<\epsilon$. We divide $[0,1]$ into segments $[t_i,t_{i+1}]$ where $0=t_0<t_1<\ldots <t_n=1$ and choose $n$ large enough so $t_{i+1}-t_i<\delta$. By the uniform continuity of $\Gamma$, if $y\in[t_i,t_{i+1}] $ then $\Gamma(y)\in B_{\epsilon}(\Gamma(t_i))\subset\Gamma^{\epsilon}\subset U$. $B_{\epsilon}(\Gamma(t_i))$ denotes the ball with radius $\epsilon$ and center at $\Gamma(t_i)$.
Now we construct a suitable polygon $\Delta:[0,1]\to\Gamma^{\epsilon} $ that satisfies our requirement. Let $i\in\{1,\ldots,n-1\}$, we set $\Delta(t_i)=\Gamma(t_i)$ - verteces of the polygon and $E_i=\Delta(t)$ be a parametrization of the edge connecting $\Delta(t_{i}), \Delta(t_{i+1})$ for $t\in[t_i,t_{i+1}] $. Then $\Delta(t_{i-1}), \Delta(t_{i+1})\in B_{\epsilon}(\Gamma(t_i))$ $(\Delta(t_{n})=\Delta(t_{0})$, i.e. $\Delta$ starts and ends at the same point) and also the corresponding edges $E_{i-1},E_i$ are subsets of the same ball. For $t\in[t_i,t_{i+1}] $ we define homotopy $\gamma(s,t)=(1-s)\Gamma(t)+s\Delta(t)$. Since $B_{\epsilon}(\Gamma(t_i))$ is convex, function $\gamma$ attains its values within this ball. Hence $\gamma:[0,1]\times[0,1]\to U$ is well defined and continuous.
I would like to know if this proof is correct or not. Thank you for any help.
This community wiki solution is intended to clear the question from the unanswered queue.
Yes, your proof is correct.