I just came across the notion of [ends](https://en.wikipedia.org/wiki/End_(topology)) of a space, and I wonder if the following are equivalent for $G$ a locally finite connected graph:


1. There exists an infinite path $v\_1,v\_2,\dots$ in $G$ which hits every vertex at least once but not infinitely many times
2. $G$ is 1-ended



Characterizing 1-ended graphs
I'll prove the direction that 2 implies 1.


Using that $G$ is connected and 1-ended, you can obtain an increasing sequence of connected, finite diameter subgraphs
$$G'\_0 \subset G'\_1 \subset G'\_2 \subset \cdots
$$
whose union is $G$, such that for all $i$, all but one component of the subgraph $G \setminus G'\_i$ has finite diameter. 


Since $G$ is locally finite, it follows that each $G'\_i$ is finite, each finite diameter component of $G \setminus G'\_i$ is finite, and there are only finitely many such components. Letting $G\_i$ be the union of $G'\_i$ with the finite diameter components of $G \setminus G'\_i$, we obtain an increasing sequence of connected, **finite** subgraphs
$$G\_0 \subset G\_1 \subset G\_2 \subset \cdots
$$
whose union is $G$, such that $G \setminus G\_i$ consists of a single unbounded component. 


Let $P\_i = G\_i \cap (G \setminus G\_i)$, which is a finite set of points. For convenience, I'll pass to a subsequence so that $P\_i$ is contained in the topological interior of $G\_{i+1}$, hence $P\_i$ is disjoint from $G \setminus G\_{i+1}$.


Now I'll construct the desired infinite path $\gamma$, inductively constructing a strictly increasing family of initial segments 
$$\gamma\_0 \subset \gamma\_1 \subset \gamma\_2 \subset \cdots
$$
The path $\gamma\_0$ starts anywhere, then runs around, concatenating edges, and hitting every vertex of $G\_0$, ending at some $p\_0 \in P\_0$. Assuming by induction that $\gamma\_i$ has been constructed to hit every point of $G\_i$, ending at some $p\_i \in P\_i$, starting from $p\_i$ the path $\gamma\_{i+1} \setminus \gamma\_i$ runs around in $G \setminus G\_i$, concatenating edges, and hitting every point of $G\_{i+1} \setminus G\_i$, ending at some point $p\_{i+1} \in P\_{i+1}$.


The path $\gamma$ does hit every vertex, by construction. Also, clearly $\gamma\_{i+1}$ hits each vertex of $G\_i$ only finitely many times and $\gamma\_{i+2} \setminus \gamma\_{i+1}$ does not hit any vertex of $G\_i$, hence $\gamma$ hits each vertex only finitely many times.


