Here is an ad hoc attempt. A connectivity space is an ordered pair $\ \mathbf X:=(X\ \mathcal C)\ $ such that the following two axioms hold:
- $\ \left(A\ne\emptyset\ne B\,\ \wedge\ \,A\ B\in\mathcal C\right)\\ \quad \Rightarrow\quad \left( A\cup B\in \mathcal C\ \ \Leftrightarrow\quad\exists_{x\in A\cup B} \left(A\cup\{x\}\in\mathcal C\ \ \wedge\ \ B\cup\{x\}\in \mathcal C\right)\ \right) $
- $\ \forall_{x\ y\,\in\,A}\ \exists_{S\in\mathcal C}\ (x\ y\in S\ \ \wedge\ \ S\subseteq A)\quad\Rightarrow\quad A\in \mathcal C$
for every $\ A\ B\ \subseteq X.\ $ Next, given connectivity spaces $\ \mathbf X:=(X\ \mathcal C)\ $ and $\ \mathbf Y:=(Y\ \mathcal D),\ $ A connectivity map (or connectivity morphism) is any function $\ f:X\rightarrow Y\ $ such that $\ \forall_{A\in\mathcal C}\ f(A)\in\mathcal D.$
Given a topological space $\ \mathbf X:=(X\ T),\ $ we get the induced connectivity space $\ \mathbf X_c := (X\ \mathcal C_T),\ $ where $\ C_T\ $ is the family of all connected subsets of $\ \mathbf X$. Thus every continuous map between two topological spaces is a connectivity map between the induced connectivity spaces.
Do not expect that there is a very close relation between continuous maps and connectivity maps. But the relation between them should be interesting (a source of new MO-questions :-) ).
REMARK It follows from the above definition (two axioms) that $\ \emptyset\in\mathcal C.\ $ (Thank you Eric for this point).
EXAMPLE An intersection $\ \bigcap_{n=1}^\infty A_n\ $of a monotone sequence of closed connected subspaces $\ A_n\ $ doesn't have to be connected. For instance, consider the following subspaces of $\ \mathbb R^2$:
$$A_n\ :=\ \mathbb R^2\setminus (-1;1)\times(-n;n)$$
for $\ n=1\ 2\ \ldots$
Of course a small modification will give a similar example for open connected subsets $\ B_n,\ $ say:
$$B_n\ :=\ \mathbb R^2\setminus \{0\}\times[-n;n]$$