[I'll ask this in the one-dimensional setting, and (if the answer is yes) I'll leave as open what possible extensions to more general settings one might wish to discuss.]
Let $K \subset \mathbb{R}$ be a compact nowhere dense set. Suppose that for each $x \in K$ we have a sequence $(U_n^x)_{n \in \mathbb{N}}$ of connected neighbourhoods of $x$ such that the length of $U_n^x$ tends to $0$ as $n \to \infty$.
Does there necessarily exist a finite set $S \subset K$ and a list of integers $(n_x)_{x \in S}$ such that $K \subset \bigcup_{x \in S} U_{n_x}^x\,$ and for all distinct $x,y \in S$, $\,U_{n_x}^x \cap U_{n_y}^y = \emptyset$?
If not, what about if we add the assumption that $K$ is a Lebesgue-null set?
(My vague intuition on the last bit is that requiring $K$ to be a null set won't make a difference to the answer, as there's probably some homeomorphic or "nearly homeomorphic" transformation of $\mathbb{R}$ that will turn a positive-measure nowhere dense set into a null set.)
If the answer is yes (with or without the null set requirement), is there a reference for this? Even just an exercise from a textbook to prove this, or prove something that easily implies this, would suffice.
The answer is no, even if $K$ is countable.
For example, let $K = \{0\} \cup \{1/k: k \in \mathbb N\}$. Let $U^0_n = (-1/n, 1/n)$, and choose $U^{1/k}_n$ so that $U^{1/k}_n \cap K = \{1/k\}$. In order for $0$ to be covered you need $0 \in S$, and $U^0_n \cap U^{1/n}_m \ne \emptyset$ for all $m$ and $n$.