Simply stated, I've been trying for a long time to either find in the literature, or derive myself, a notion of path in Cech closure spaces, that specialises to paths in a topological space, and to graph-like paths in so-called "quasi-discrete closure spaces".

Let me recall the definitions:

A closure space is a pair $(X,C)$ where $C : \mathcal P (X) \to \mathcal P (X)$ is a function satisfying $C(\emptyset) = \emptyset$, $A \subseteq C(A)$, $C(A \cup B) = C(A) \cup C(B)$.

A continuous function $f$ is a function between two spaces such that $f(C(A)) \subseteq C(f(A))$

A topological space is (via the Kuratowski definition) a closure space with the additional axiom $C(C(A)) = C(A)$ (idempotence of closure).

Any reflexive relation $R$ generates a closure space by $C(A) = \{y \in A | \exists x \in A . x R y\}$. That's called a "quasi-discrete closure space".

Topological paths are defined as continuous functions from the unit interval.

Let me now make two examples.

Example 1: $\mathbb R^2$. Topological paths work fine (indeed!).

Example 2: the closure space on $\mathbb N$ generated by the successor relation. It's a nice closure space, but topological paths exist that do not "follow the edges" of non-symmetric relations, due to "directionality" of $R$; topology (e.g. the unit interval) is intrinsically symmetric; relations are not. For an example of this consider the set $\{a,b\}$ and the relation $R = \{ (a,b) \}$. This generates a quasi-discrete closure space. Consider the function $f : [0,1] \to \{a,b\}$ with $f(0) = b$ and $f((0,1])) = a$. This function is continuous but not a graph-like path in $R$.