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 $\epsilon >0$.