For Cantor-connectedness I use the following definition:
A $p$-metric space $(X,d)$ is Cantor-connected if for any $\epsilon > 0$, any two points $x, y \in X$ can be connected by an $\epsilon$-chain, i.e., there exist points $x = x_{0}, x_{1}, \cdots, x_{n} = y$ in $X$ such that $d(x_{i-1},x_{i}) \leq \epsilon$ for all $i \leq n$.
I would like to prove the following characterisation of Cantor-connectedness:
A space $(X,d)$ is Cantor-connected if and only if it cannot be partitioned into two sets $A$ and $B$ such that $d(A,B) > 0.$
I already have the following:
We first prove that if $(X,d)$ is Cantor-connected, $X$ cannot be partitioned into two sets $A$ and $B$ such that $d(A,B) > 0.$ Suppose that there exists a partition of $X$ into sets $A$ and $B$ with $d(A,B) > 0.$ Take $a \in A$ and $b \in B$, take $\epsilon > 0$. By the fact that $X$ is Cantor-connected, there exist $a = x_{0}, x_{1}, \cdots, x_{n} = b$ in $X$ such that $d(x_{i-1},x_{i}) \leq \epsilon$ for all $i \in \{1, \cdots,n\}$. But then we get $d(a,b) \leq \sum_{i=1}^{n}d(x_{i-1},x_{i}) \leq n \epsilon.$ By arbitrariness of $\epsilon$ we get that $d(a,b) = 0$, hence $d(A,B)=0$. This is a contradiction.
Now we show that if $X$ cannot be partitioned into two sets $A$ and $B$ with $d(A,B) > 0$, then $X$ is Cantor-connected. Suppose that $x, y \in X$ and $\epsilon > 0$ arbitrary. Set $x_{0} = x$. Then there exists $x_{1} \in X$ such that $d(x_{0},x_{1}) \leq \epsilon$. If not, $\{x\}$ and $X \setminus \{x\}$ would form a partition with $d(\{x\}, X \setminus \{x\}) > 0$. Analogously we find $x_{2} \in X$ such that $d(x_{1},x_{2}) \leq \epsilon.$ This will give us $x_{0}, x_{1}, x_{2}, \cdots \in X$ with $d(x_{i-1},x_{i}) \leq \epsilon.$ Now we have to prove that after a finite number of steps, we get $x_{n}= y$ and $d(x_{n-1}, x_{n}) \leq \epsilon.$
Can anyone explain to me why the process in the last step stops after a finite number of steps?