Does $(\omega, E)$ with the cycle condition have an $\omega$-path?

Question

Let $G = (V,E)$ be a simple, undirected graph. We say that $v\neq w\in V$ lie on a common cycle if there is an integer $n\geq 3$ and an injective graph homomorphism $f: C_n\to V$ such that $v,w\in \text{im}(f)$.

We call a function $p:\omega\to V(G)$ an $\omega$-walk if for all $n\in \omega$ we have $\{p(n),p(n+1)\}\in E(G)$.

Let $(\omega,E)$ be a graph on the ground set $\omega$ such that for all $m\neq n\in \omega$ we have that $m$ and $n$ lie on a common cycle.

Questions.

1. Is there necessarily an injective $\omega$-walk (also called an $\omega$-path) for $(\omega,E)$?
2. If not, is there an $\omega$-walk $p:\omega\to\omega$ such that $p^{-1}(\{n\})$ is finite for all $n\in \omega$?

Answer 1

The answer to both questions in negative.

Let $G$ be the graph on $\omega$ with $nE0$ and $nE1$ for all $n\geq 2$ (and no other edges). Clearly all $m\neq n$ with $n,m\geq 2$ are on a cycle of length $4$, given by $nE0EmE1En$. However any $\omega$-walk must go through either $0$ or $1$ infinitely many times.

Answer 2

No. Draw edges from $0$ and $1$ to all numbers $n>1$. Now any two nodes lies on a cycle of length $4$. But there is no injective $\omega$-walk, since every edge touches either $0$ or $1$, and indeed every walk must use one of these nodes infinitely often.