Let $U\subseteq\mathbb{R}^n$ be an open, connected set, and let $a,b\in U$. Is it always true that there exists open balls $B_1,\ldots,B_k\subseteq U$ such that $a\in B_1,b\in B_k$, and $B_i\cap B_{i+1}\neq\emptyset$?
I think it should be true, but it's not clear how this will follow from the definition of a connected set (cannot be partitioned into two nonempty disjoint open sets.)
Pick $a\in U$ and let $A$ be the set of points for which suc a ball chain to $a$ exists. Then $A$ is open as from any reachable point $x$ we may reach all points in a ball around $x$. But $A$ is also (relatively) closed because if $x$ is not reachable, then also points sufficiently close to $x$ are not reachable (namely if $B_r(x)\subseteq U$, then this holds for all points in $B_{\frac r2}(x)$). The only nonempty clopen subset of a connected space is the whoile space, voila.
This is essentially the usual argument to show path-connectedness of connected opne subsets of $\mathbb R^2$, but here applied directly to "ball-chain" connectedness.
It is because connected open set in $\mathbb R^n$ is path connected: see this post:Connectivity, Path Connectivity and Differentiability. Using this you can form a path joining $a$ and $b$, and use finitely many open balls to cover this path. These balls would satisfy your condition.
