This question was inspired to me by Lukas Geyer’s recent question.
A positive answer to this question would easily entail a positive answer to Lukas’ question also, and a negative answer would probably be informative as well.
Let $T=]0,1[^2$ be the open unit square. Let $(S_k)_{k\geq 1}$ be a countable family of open segments in $T$. What we call a cycle is a finite sequence $M_1,M_2, \ldots ,M_r$ of distinct points in $T$, not all on a line (thus $r>2$) such that all the segments $]M_1M_2[,]M_2M_3[, \ldots ,]M_{r-1}M_r[$, and also $]M_rM_1[$, are each included in one of the segments $S_k$, and likewise all the endpoints $M_1,M_2, \ldots ,M_r$ are each in one of the $S_k$. Formally, for each $i\in{\mathbb Z}/r{\mathbb Z}$ there is an index $k(i)$ with $]M_iM_{i+1}[ \subseteq S_{k(i)}$ and there is an index $l(i)$ such that $M_i \in l(i)$.
What we call a cut is a finite sequence $M_1,M_2, \ldots ,M_r$ of distinct points such that the first and the last, $M_1$ and $M_r$, are on the boundary of $T$, and the intermediate points $M_2,M_3, \ldots ,M_{r-1}$ are in $T$, and all the segments $]M_1M_2[,]M_2M_3[, \ldots ,]M_{r-2}M_{r-1}[, ]M_{r-1}M_r[$, are each included in one of the segments $S_k$, and likewise all the intermediate points $M_2,M_3, \ldots ,M_{r-1}$ are each in one of the $S_k$.
So the question is as follows : assume that there are no cycles or cuts. Does it follow that
$ Z=T \setminus \bigg(\bigcup_{k\geq 1} S_k \bigg) $
is necessarily path-connected ?
Update 16:15 : The definitions have been corrected so as to avoid the uninteresting counterexample explained in Dan Shved’s answer.
Update 19:10 : Dan Shved’s second counterexample strikes a much stronger blow to my hopes, because it shows a family $(S_k)$ such that $T \setminus (\cup_{k \geq 1} S_k)$ is not path-connected, and yet $T\setminus (\cup_{1 \leq k \leq n} S_k)$ is for every finite $n$. So there is no “finite” sufficient condition.