Let $(X, \mathcal{U})$ be a uniform space. We know that every uniform space $(X, \mathcal{U})$ becomes a topological space $(X, \tau_{\mathcal{U}})$ by defining a subset $G$ of $X$ to be open, $G\in \tau_{\mathcal{U}}$, if and only if for every $x$ in $G$ there exists an $U\in \mathcal{U}$ such that $U[x]$ is a subset of $G$.
The non-empty subset $A\subset X$ is called chain connected if for every $x,y\in A\subseteq X$ and every $U\in\mathcal{U}$,there is $U$-chain connected from a point $x\in X$ to a point $y\in X$ in uniform space $(X,\mathcal{U})$, this means that there is a finite sequence of points $x=x_l, x_2,...,x_n=y$ such that $(x_i, x_{i+l})\in U$ for each $i=1,2 .... ,n-l$.
Question. Is it true that $A$ is chain connected if and only if $A$ is connected?
No, in $\mathbb{R}$ in the standard uniformity induced by $d(x,y) = |x-y|$, $A = \mathbb{Q}$ is "chain-connected" but not connected. A connected set is always chain-connected (by the chain characterisation of connectedness using open covers).
I think (IIRC) chain-connected can be characterised as : "every uniformly continuous function $f: A \to \{0,1\}$, where the latter space has the discrete uniformity, is constant”. It's also called "uniformly connected".