This is a follow-up to another question concerning infinite paths which was admittedly ill-posed. I hope this question is posed better.
The graph $N$ with vertex set $V(N) = \mathbb{N}$ and $(x,y) \in E(N)$ iff $x < y$ contains paths of arbitrary length connecting 0 and an appropriate $n$ .
The graph $Q$ with vertex set $V(Q) = \mathbb{Q}$ and $(x,y) \in E(Q)$ iff $x < y$ contains paths of arbitrary length connecting 0 and 1.
Of course, both graphs contain infinite paths, starting from 0, but ending nowhere.
It's more or less obvious, that $N$ doesn't contain a path of infinite length connecting 0 and an $n\in \mathbb{N}$ (because all $n$ are finite).
But it's hard (for me) to "see" and get a feeling why $Q$ doesn't contain a path of infinite length connecting 0 and 1: each finite path between 0 and 1 is a finite subset of $E(Q)$: $\lbrace (0,q_1),(q_1,q_2),...,(q_n,1)\rbrace$. Why cannot there be an infinite subset of that kind?
Is it impossible to define "that kind", i.e. the property "being a path connecting 0 and 1", or can we define it (in second order logic maybe) but prove, that no infinite subset of $E(Q)$ with this property exists?