I'm trying to prove the following.
"Let $\Omega \subseteq \Bbb{C}$ be a domain. and let $\epsilon>0$. For any two points $u,z\in\Omega$, prove that there exists a sequence of points ${w_i}^{N}_{i=1}\subseteq\Omega$ satisfying $w_1=u$, $w_N=z$, and $|w_i -w_{i+1}|<\epsilon$ for all $i=1,...,N-1$."
It is obvious to show that domain is path-connected by setting two disjoint open subsets and concatenate the path within the points to show that one of the open subsets is, in fact, the whole domain. However, I'm not fully understanding how to go by solving the actual proof where we are showing the two consecutive points have distance less than some $\epsilon >0$
edit - Sorry I forgot to mention... I'm not supposed to use the path-connectedness
A typical argument to show that all elements of a connected set $\Omega$ satisfy a certain property $P$ is to show that both the subset of all points satisfying $P$ and its complement are open in $\Omega$. This works here as well:
For fixed $u \in \Omega$ and $\epsilon > 0$ define two sets $$ A = \{ z \in \Omega \mid \text{There exists a chain } u = w_1,w_2, \ldots w_{N-1}, w_N = z \text{ in } \Omega \\\text{ with } |w_i -w_{i+1}|<\epsilon \text{ for } 1 \le i \le N-1 \} $$ and its complement $B = \Omega \setminus A$.
Then show that both $A$ and $B$ are open sets: If $z \in A$ (resp. $B$) and $U_\delta(z) \subset \Omega$ for some $0 < \delta \le \epsilon$, then $U_\delta(z) \subset A$ (resp. $B$). ($U_\delta(z)$ denotes the open disc with center $z$ and radius $\delta$.)
Since $\Omega$ is connected, one of them must be empty, and consequently, $\Omega = A$.