Countable graph that is "as non-traceable as it gets"

Question

If $\omega$ denotes the set of the natural numbers (= the first infinite ordinal), and if $E\subseteq\binom{\omega}{2}$ is any subset, we call a map $f\colon\omega\to\omega$ a walk in the graph $(\omega,E)$ if and only if $\{f(k),f(k+1)\}\in E$ for all $k\in\omega$.

We call $f$ a walk a Hamiltonian walk¹ if and only if it is surjective.

Question.

Is there $E \subseteq \binom{\omega}{2}$ such that

1. there is at least one Hamiltonian walk in $(\omega,E)$,
2. every Hamiltonian walk in $(\omega,E)$ visits every vertex infinitely-often?

Remarks.

• _{A walk need not be injective, nor need it be surjective.}

• _{Condition 1. implies that $(\omega,E)$ is a connected graph.}

• _{In graph theory, an injective walk is called a path. (This is accidentally different from the usual convention in topology, where the term 'path' signals that self-intersections are permitted.)}

• _{In graph theory, a graph admitting an injective and surjective walk into it is called traceable. (For finite graphs, the term Hamiltonian most often means that there exists a Hamilton path whose end-vertices are adjacent, i.e., a Hamilton circuit. For infinite graphs, this does not make sense (simply because then a Hamilton path does not have two "end-vertices"), which necessitates either (0) keeping to the study of Hamilton-paths only, or (1) using definitions involving some kind of 'convergence'.) This explains the focus of this question on Hamilton-paths, and the title of the OP.}

¹_{This terminology harmonizes rather well with e.g. the book Futaba Fujie, Ping Zhang: Covering Walks in Graphs. SpringerBriefs in Mathematics, 2014.}