For a metric space $X$, show that if $F \subset X$ is closed and connected then for every pair of points $a, b \in F$ and each $\varepsilon > 0$ there are points $z_0, z_1, ... , z_n \in F$ with $z_0 = a$, $z_n = b$ and $d(z_{k−1},z_k) < \varepsilon$ for
$1 \le k \le n$. Is the hypothesis that $F$ be closed needed?
Actually, I tried to prove it by contradiction, i.e. let there exists distinct $a, b \in F$ and there does not exist any finite set of points in F such that the above condition holds . Then, $\exists \varepsilon>0$ such that for any finite set of points $\{a, z_0, z_1, ..., z_n\} \in F$, there is a positive integer $k<n$ with $d(z_{k−1},z_k) > \varepsilon$ , then for any $z \in F$, $z \ne a,b$ ; either $d(a,z) < \varepsilon$ or $d(a,z) \ge \varepsilon$, now as $F$ is connected, there is a $z$ with $d(a,z) = d(b,z) = \varepsilon$, then I can't approach .
We don’t need the assumption that the set $F$ is closed, because we can restrict our consideration to the case when $X=F$ (and is non-empty). Fix any $\varepsilon>0$. Define an equivalence relation $\sim$ on the set $F$ as follows. For each $x,y\in F$ put $x\sim y$ iff there exist $z_0, z_1, ... , z_n \in F$ with $z_0 = x$, $z_n = y$ and $d(z_{k−1},z_k) < \varepsilon$ for $1 \le k \le n$. Let $K$ be any equivalence class of the relation $\sim$ and $x\in K$ be any point. If $y\in F$ and $d(x,y)<\varepsilon$ then $y\in K$, so the set $K$ is open in $F$. Since the partition of the set $F$ into the equivalence classes is a partition into open (in $F$) sets, each of these classes is also closed. But, since the set $F$ is connected, this is possible only when there is only one equivalence class. This means that any two points $a,b\in F$ are equivalent, that is the condition holds.