Alternative characterization of connectivity spaces. First, a barrage of definitions.
Define that a relation on a set $X$ is a subset of $X^2$. If $f$ is a symmetric relation on $X$, define that $f$ is connected if and only if there do not exist disjoint and exhaustive $A, B \subseteq X$ and non-empty relations $\alpha,\beta$ on $A$ and $B$ respectively such that $f = \alpha \cup \beta$.
Given a collection of sets $\mathcal{J}$, define the that overlap relation is the set $$\mathcal{J}^{\mathrm{ov}} = \{(A,B)|A,B \in \mathcal{J}, A \cap B \neq \emptyset\}.$$
Finally, define that a connectivity space is a pair $(X,\mathcal{K})$ with $\mathcal{K} \subseteq \mathcal{P}(X)$ such that


*

*$\emptyset \in \mathcal{K}$

*For all non-empty $\mathcal{J} \subseteq \mathcal{K}$ such that $\bigcap \mathcal{J} \neq \emptyset,$ it holds that $\bigcup \mathcal{J} \in \mathcal{K}$.
(Note: It is typical to call the elements of $\mathcal{K}$ "connected sets." Also, I presume that the set of all connected relations on a set $X$ forms a connectivity space; however, this is tangential to the issue.)
Question: Suppose we replaced condition 2 in the definition of a connectivity space with the following.
2.* For all non-empty $\mathcal{J} \subseteq \mathcal{K}$ such that $\mathcal{J}^{\mathrm{ov}}$ is connected, it holds that $\bigcup \mathcal{J} \in \mathcal{K}$.
Is the resulting definition equivalent to the original?
 A: First we show $(2^\ast) \rightarrow (2)$:  
Assume $(2^\ast)$, and let $\mathcal{J} \subset \mathcal{K}$ be arbitrary, but such that $\bigcap \mathcal{J} \neq \emptyset$.  We can conclude $\bigcup \mathcal{J} \in \mathcal{K}$ if we can show that $\mathcal{J}^{ov}$ is connected.  We're assuming that $\bigcup \mathcal{J} \neq \emptyset$, so $\mathcal{J}^{ov} = \mathcal{J}^2$, i.e. the "full" relation on $\mathcal{J}$.  It's not hard to see that this relation is connected; let's extract this as a:
Lemma: if $S \neq \emptyset$ is a set, then $S^2$ is connected.
Proof: Suppose not, and let $A, B, \alpha, \beta$ witness this.  $A, B$ are non-empty, so let's pick elements from them, say $a, b$.  Clearly $(a,b) \in S^2$, but it's impossible for $(a,b) \in \alpha \cup \beta$ since $\alpha \subset A^2$, $\beta \subset B^2$, and $A, B$ are disjoint from one another.
Now $(2) \rightarrow (2^\ast)$:  
Let $\mathcal{J}$ be such that $\mathcal{J}^{ov}$ is connected.  We'll apply Zorn's Lemma to see that $\bigcup \mathcal{J}\in\mathcal{K}$.  Let 
$$\mathbb{P} = \left\{ \mathcal{I}\subset\mathcal{J}\ |\ \bigcup \mathcal{I}\in\mathcal{K}\right\}$$
And partially order $\mathbb{P}$ by inclusion.  $\mathcal{P}$ is not empty: for any $J \in \mathcal{J}$, $\mathcal{I} = \{J\}$ witnesses the non-emptiness.  Every chain in $\mathbb{P}$ has an upper bound in $\mathbb{P}$: this follows from your observation about nested collections of connected sets, which follows immediately from $(2)$.  Thus by Zorn's Lemma $\mathbb{P}$ has a maximal element.  Let's pick one and call it $\mathcal{I}$.  I claim that $\mathcal{I} = \mathcal{J}$ (and so $\bigcup\mathcal{J}\in\mathcal{K}$, as desired).  
Suppose not.  Case 1: there is some $J\in\mathcal{J}\setminus\mathcal{I}$ such that $J\cap\bigcup\mathcal{I}\neq\emptyset$.  If that were the case, then by $(2)$, $\{J\}\cup\bigcup\mathcal{I}\in\mathcal{K}$, but then $\{J\}\cup\mathcal{I}\in\mathbb{P}$, contradicting the maximality of $\mathcal{I}$.  Case 2: Every $J\in\mathcal{J}\setminus\mathcal{I}$ is disjoint from $\bigcup\mathcal{I}$.  This also yields a contradiction, for $A=\mathcal{I}$, $B=\mathcal{J}\setminus\mathcal{I}$ witness that $J^{ov}$ is disconnected.
