The question is the same as the title. Precisely,
Suppose $\Omega \subseteq \mathbb{C}$ is an open connected domain. Given two points $z_0, z_1 \in \Omega$, does there exist a rectifiable curve $\gamma : [0, 1] \to \Omega$ with $\gamma(0) = z_0, \gamma(1) = z_1$. What about piecewise $C^1$ $\gamma$ ?
Since $\mathbb{C}$ is locally path connected and $\Omega$ is connected, we have that $\Omega$ is path connected, so that there is a continuous curve $\gamma$ which starts at $z_0$ and ends at $z_1$. But can we always choose $\gamma$ to be rectifiable or piecewise $C^1$ (the latter would imply the former though).
Best Answer
Yes, we can. Take some continuous curve $C$ connecting the two points. Choose some open covering $\mathcal U$ of the curve by discs whose center is on the curve and which are contained in $\Omega$. Since the curve is the continuous image of a compact interval, it is itself compact, and thus there exists a finite subcovering of the curve by discs centered on the curve. For simplicity, add two additional discs, one centered at the start point, one at the end point.
Now connect the centers of intersecting discs by straight lines. You get a connected graph whose (finite number of) edges are contained in $\Omega$, and connect start and end point. Now just choose any path from the start to end point along these edges.
To be completely rigorous, you need to fill in some blanks, though: Why can we choose $\mathcal U$ as desired? Why is the graph connected? Why are the edges contained in $\Omega$?