TITLE: Characterization of finite covering spaces, and the existence of a non-finite analogue
QUESTION [9 upvotes]: For this post I want to assume that $X$ is connected, Hausdorff, locally compact, and semi-locally simply connected topological space but, honestly, I’d be happy to assume that $X$ is a topological manifold (or even a complex manifold).
There is a well-known result which characterizes when a map $f\colon Y\to X$ (where $Y$ is also Hausdorff) is a finite covering space in more generally used topological terms.

Fact (e.g. see [Ho, Lemma 2]): The map $f\colon Y\to X$ is a finite covering space if and only if the following two conditions
hold:

$f$ is a local homeomorphism,
$f$ is proper (*).


I am interested in understanding whether or not (again perhaps with stronger assumptions on $X$) there exists a version of this fact with ‘finite covering space’ replaced by ‘covering space’. Or, perhaps more realistically (since I doubt such a general characterization exists), a more general class of covering spaces which can be characterized in terms similar to this fact.
Of course, what needs to be relaxed is that $f$ is proper. For reasons related to my motivation from algebraic geometry, I would like to replace “$f$ is proper” with “$f$ is $P$” where, intuitively, one has an ‘equality’
$$P=\text{proper}-“\text{quasi-compact}”.$$
For instance, by the contents of Tag 005M (and more explicitly Tag 005R) one has that
$$\text{proper}=\text{quasi-proper}+\text{closed},$$
where one defines a map $f\colon Y\to X$ to be quasi-proper if $f^{-1}(V)$ is quasi-compact for all quasi-compact subsets $V\subseteq X$ (note that since my spaces are going to generally be Hausdorff one can replace ‘quasi-compact’ with ‘compact’). Thus, one might try to replace “proper” with “closed” in the statement of this fact. Of course, “closed” is a very poor choice. For one thing it doesn’t even include the example of the universal covering $\mathbb{R}\to S^1$. Moreover, “closed” doesn’t even jive with what I am intuitively going for. But maybe this gives an indication of the sort of direction I’d be interested in going.
Any suggestions would be greatly appreciated!
(*) Here by proper I mean universally closed as in Tag 005O, but I think that one can do away with the assumption that $X$ (and $Y$) are Hausdorff by using a slight modification which is more aligned with my algebro-geometric background. Indeed, it might be more correct to write ‘proper’ to mean ‘universally closed and separated’ (where separated is as in Tag 0CY0). Of course, if $Y$ and $X$ are Hausdorff, separatedness is automatic.
References:
[Ho] Ho, C.W., 1975. A note on proper maps. Proceedings of the American Mathematical Society, 51(1), pp.237-241. https://www.ams.org/journals/proc/1975-051-01/S0002-9939-1975-0370471-3/S0002-9939-1975-0370471-3.pdf

REPLY [2 votes]: Suppose that $f: X\to Y$ is a local homeomorphism of reasonably nice  topological spaces: Throughout, I will assume that both are metrizable, connected, locally path-connected and semilocally simply-connected (although this is an overkill for most of the arguments).
Here is the key ingredient that, according to your comment, you already knew how to prove once upon a time:
Lemma 1. Suppose that $c: [0,1]\to X$ is a path that has a partial lift
$\tilde c: [0,1)\to Y$, such that $\tilde c([0,1))$ is relatively compact in $Y$. Then the lift $\tilde c$ extends continuously to the entire interval $[0,1]$.
Proof. Let $K$ be the compact closure of  $\tilde c([0,1))$ in $Y$. Take a sequence $t_i\in [0,1)$ converging to $1$ such that
$(\tilde{c}(t_i))$ converges to some $y\in K$. Then set
$$
\tilde{c}(1):= y.
$$
Let's prove that $\tilde{c}: [0,1]\to Y$ defined this way is continuous. For the sake of a contradiction, let $s_i\in [0,1)$ be a sequence such that $(\tilde{c}(s_i))$ converges to some $z\in K$, $z\ne y$. Find neighborhoods $U, V$ of $y, z$ in $Y$ such that $f|_U, f|_V$ are homeomorphisms to a common neighborhood $W$ of $x=c(1)$. WLOG, we can assume that $U, V$ are disjoint and that both sequences $(t_i), (s_i)$ belong to $c^{-1}(W)$. But then for, $s_i< t_i$, the image
$$
\tilde c([s_i, t_i])
$$
cannot intersect $f^{-1}(\partial W)$. Hence, that image is entirely contained in both $U$ and in $V$, which is a contradiction. qed
Definition 1. A  map $f: X\to Y$ is locally proper if for every $y\in Y$ there is a neighborhood $V$ of $y$ in $Y$ such that
whenever $c: I=[0,1]\to V$ is a continuous map, each path-connected component of $f^{-1}(c(I))$ is relatively compact in $X$.
Corollary 1. If $f: X\to Y$ is a locally proper local homeomorphism, then it satisfies the path-lifting property, hence, is a covering map.
Proof. The proof is the standard reduction of the path-lifting property to the extension property for partial lifts $\tilde{c}: [0,1)\to Y$. Namely,
given a path $c: [0,1]\to Y$ and $x\in f^{-1}(c(0))$, you find a maximal subinterval $J$ in $I=[0,1]$ containing $0$ such that $c|_J$ lifts to a path $\tilde{c}: J\to Y$, such that $\tilde c(0)=x$.
Since $f$ is a local homeomorphism, $J$ is an open subset of $I$, hence, has the form $[0,T)$, unless it is $[0,1]$ in which case we would be done. Let $y=c(T)$ and let $V$ be a neighborhood of $y$ as in Definition 1. By continuity, there is a subinterval $[S,T]\subset I$ such that $c([S,T])\subset V$. Now, let $A$ be the path-connected component of $f^{-1}(c([S,T]))$ containing $\tilde{c}([S,T))$. By the assumption, it is relatively compact. Now, apply  Lemma 1, to conclude that $\tilde{c}$ extends continuously to $T$. qed
Remark. If $f$ is a covering map, then it is quite immediate that it satisfies the local properness condition: Just take a neighborhood $V$  of $y$ such that $f^{-1}(V)$ breaks into open subsets $U_\alpha$ such that each restriction $f|_{U_\alpha}: U_\alpha\to V$ is a homeomorphism.
Now, maybe you are already happy with Definition 1. If not, here are two more definitions (names, as before, are completely made-up).
Recall that a topological space $A$ is called an arc if it is homeomorphic to $[0,1]$. A Peano continuum is a metrizable space which is a continuous image of $[0,1]$. Equivalently, it is a connected, locally connected,  compact metrizable space.
Definition 2. A continuous map $f: X\to Y$ is arc-proper if for each  arc $A\subset Y$, each component of $f^{-1}(A)$ is compact.
Definition 3. A continuous map $f: X\to Y$ is simply-proper if for each Peano continuum $C\subset Y$ for which the induced map $\pi_1(C)\to \pi_1(Y)$ is trivial, each component of $f^{-1}(C)$ is compact.
The last definition might be the most appropriate for algebro-geometric purposes.
Conjecture. Suppose that $f: X\to Y$ is a local homeomorphism as before. Then the following are equivalent:

$f$ is a covering map.

$f$ is simply-proper.

$f$ is arc-proper.


Edit 1.
The implications 1 $\Rightarrow$ 2  $\Rightarrow$ 3 are quite clear, I think, the remaining implication 3  $\Rightarrow$ 1 also holds (see below).
Lemma 2. Suppose that $f: X\to Y$ is a local homeomorphism as before. Then the following are equivalent:

$f$ is a covering map.

$f$ is simply-proper.


Proof. This is a direct consequence of the assumption that $Y$ is semilocally simply-connected and Lemma 1. qed
Lastly,  I will prove the implication 3  $\Rightarrow$ 1 in the setting of smooth maps between smooth manifolds
(which you seem to be happy with):
Lemma 3. Suppose that $X, Y$ are connected smooth manifolds, $f: X\to Y$ is an arc-proper local diffeomorphism. Then $f$ is a covering map.
Proof. I will use yet another characterization of covering maps which you can find, for instance, in volume 1 of Kobayashi--Nomizu's book "Foundations of Differential Geometry":
Theorem. A local diffeomorphism between smooth manifolds  $f: X\to Y$ is a covering map if and only if for every complete Riemannian metric $g$ on $Y$, the pull-back metric $f^*g$ is again complete.
Recall that completeness of a Riemannian metric is equivalent to the property that each geodesic interval $c: [a,b]\to (M,g)$ extends to a geodesic ray.  Thus, let $g$ be a complete metric on $Y$ and assume, for the sake of a contradiction, that $h:= f^*g$ is an incomplete metric. Thus, there exists a geodesic  $c: [0, T)\to (X,h)$ which does not extend to $T$. It's easy to see that this implies properness of the map $c: [0,T)\to X$. Take a sequence $t_i\to T-$. Then $f(c(t_i))$ is a Cauchy sequence in $(Y,g)$, hence (by completeness of $g$), converges to some $y$. This the Cauchy property  holds for any sequence $t_i$ converging to $T$, it follows that
$$
\lim_{t\to T-} f\circ c(t)=y. 
$$
Let $U$ be a convex neighborhood of $y$ in $(Y,g)$ (a neighborhood where any two points are connected by a geodesic segment and this segment is unique). Then for sufficiently large $k$, $fc([t_k, T))$ is contained in $U$ and the composition map $f\circ c: [t_k, T)\to U$ is 1-1. It follows that  $A:=f\circ c([t_k, T])$ is a (necessarily geodesic, even though we do not need this) geodesic segment in $(Y,g)$, in particular, an arc. At the same time, $c([t_k, T))$ is a noncompact component of  the preimage $f^{-1}(A)$. This contradicts the assumption that $f$ is arc-proper. qed
Edit 2. I can now prove the above conjecture that arc-properness of a local homeomorphism implies the covering property. The proof is a bit tricky and is more geometric than topological, using geodesic metrization of Peano continua (Menger's conjecture, proven independently by Bing and Moise). I will refrain from writing it down since it's unclear if it is of any use.