Suppose that $L\subseteq {\cal P}(\omega)$ has the following properties:
It is not hard to see that $L$ is countable. Is there a bijection $p:\omega\to L$ such that for all $n\in\omega$ we have $p(n)\cap p(n+1) \neq \emptyset$?
Clearly, $L$ is countably infinite. Thus it will suffice to prove the following lemma, which shows that any finite path $\langle e_1,\dots,e_k,a\rangle$ in $L$ can be extended to include any new element $b\in L.$
Lemma. Given a finite set $E=\{e_1,\dots,e_k\}\subseteq L$ and $a,b\in L\setminus E,$ we can find $c,d\in L\setminus E$ (not necessarily distinct) such that $a\cap c,\ c\cap d,$ and $d\cap b$ are nonempty.
Proof. We consider two cases.
Case 1. All elements of $E$ are finite.
Choose $m\in a,\ n\in b,$ and $p\in\omega\setminus\bigcup E.$ Find $c,d\in L$ such that $\{m,p\}\subseteq c$ and $\{n,p\}\subseteq d.$
Case 2. There is an infinite element $u\in E.$
Choose $m\in a\setminus u,\ n\in b\setminus u,$ and $p\in u\setminus\bigcup(E\setminus\{u\}).$ Find $c,d\in L$ such that $\{m,p\}\subseteq c$ and $\{n,p\}\subseteq d.$
Indeed there is when L consists of finite sets. Pick an enumeration of L, and start the path with P(1)=L(1). If it comes time to pick P(n+1), see if the smallest so far unchosen member U of L intersects P(n). If it does, choose it for P(n+1). Otherwise pick m outside of all the edges chosen thus far for P and outside of U. Let P(n+1) contain m and an element of P(n) and P(n+2) contain m and an element of U. Then U is P(n+3).
This algorithm may not work if L contains an non Eulerian graph plus a set of infinite edges. The sticking point is guaranteeing the existence of m at every step. If this can be shown, then such a graph is Eulerian.
Edit 2017.02.20 GRP: It seems this needs more subtle handling, so I suggest the following modification.
As before, enumerate the set of lines $L$ and start building the path by setting $P(1)=L(1)$. Suppose after finding $n$ lines for $P$, we would like to add $U$ to the path, where $U$ is the least member of $L$ not part of $P$. If $P(n)$ and $U$ share an element in common, set $P(n+1)=U$ and continue.
If $P(n)$ and $U$ are disjoint, consider the subsets of $L$ which have none of the $n$ members of $P$ already chosen, one subset which intersects $P(n)$, and the other subset of lines which intersect $U$. If these two subsets of lines have a line $V$ in common, then set $P(n+1)=V$ and $P(n+2)=U$ and continue.
If there is no $V$ in both subsets, see if there is a point $m$ such that $m$ is on a line intersecting $P(n)$ and also on a line intersecting $U$. If there is, assign $P(n+3)=U$, and make the natural choices for $P(n+1)$ and $P(n+2)$.
After all this, one would expect $V$ or $m$ to exist, solving the problem. However, there may be the rare graph in which at some stage $n$, we cannot find $V$ or $m$. Here is where we use the intersection information to help us.
If there was one or no infinite sets in the collection , we could proceed in the finite case above, if necessary by letting $P(1)$ be the infinite set. Otherwise there are at least two infinite sets, and thus the number of lines intersecting $P(n)$ and not already in the path is infinite, as is the case for $U$ when we fail to find $V$ (and possibly fail to find $m$). Never fear, for there is a line outside of the selected path so far which connects a point that is one hop from $P(n)$ to another which is one hop from $U$. Thus $P(n+4)$ gets $U$, $P(n+2)$ gets this outside line, and pick the other two lines accordingly.
I have a hard time imagining when $m$ might not exist. This might have arisen when one of the sets of lines intersecting $U$ or intersecting $P(n)$ turns out to be empty. But both sets of lines turn out to be infinite, or at least sufficiently numerous. End Edit 2017.02.20 GRP
Gerhard "Walking In Circles Around This" Paseman, 2017.02.20.
