$F$ closed and connected then... Let $F\subset X$ be a closed and connected subset of the metric space $(X,d)$ then for every pair of points $a, b\in F$ and each $\epsilon>0$ there are points $a=z_0,z_1,\ldots z_n=b$ in $F$ such that $d(z_{k-1}-z_k)<\epsilon$ for $1\leq k\leq n$.
Is the hypothesis that $F$ is closed needed?
 A: $F$ can be any connected subset of $X$. Given $\epsilon>0$, let $\mathscr{U}=\{F\cap B(x,\epsilon/2):x\in F\}$, where $B(x,r)$ is the open $d$-ball of radius $r$ centred at $x$. Now apply the chain characterization of connectedness proved in this answer (mentioned by David Mitra in the comments) to conclude that for any $a,b\in F$ there is a finite sequence $x_0=a,x_1,\dots,x_n=b\in F$ such that $B(x_k,\epsilon/2)\cap B(x_{k+1},\epsilon/2)\ne\varnothing$ for $k=0,\dots,n-1$. Clearly $d(x_k,x_{k+1})<\epsilon/2+\epsilon/2=\epsilon$ for $k=0,\dots,n-1$.
A: You're right, this hypothesis is useless.
Let $F \subset (X,d)$ a connected subset. Let $\epsilon > 0$ and $a, b \in X$. I'm going to prove that an $\epsilon$-chain exists between $a$ and $b$. Let $\Omega \subset F$ the subset of all $\omega \in X$ such that an $\epsilon$-chain exists between $a$ and $\omega$. 
$\Omega$ is clearly open: if $\omega \in \Omega$, then $B(x,\epsilon/2) \subset \Omega$. It is also closed (in $F$): if $\omega_n$ is a sequence of elements of $\Omega$ converging to $x \in F$, one of the $\omega_n$ is at distance $< \epsilon/2$ of $x$, and that implies $x \in \Omega$. The connectedness of $F$ then implies that $\Omega$ (which is nonempty: $a \in \Omega$) is all of $F$ : $F$ is $\epsilon$-connected for all $\epsilon >0$.
