Characterization of finite covering spaces, and the existence of a non-finite analogue 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
 A: Suppose $f:X\to Y$ is a local homeomorphism and $Y$ are "locally nice" spaces. Then $f$ is a covering map if and only if $f$ uniquely lifts all paths rel. starting point. In general, local homeomorphisms which uniquely lift all paths are called semicoverings. Semicoverings have their own uses for wilder spaces but may provide a suitable bridge for you. Under the usual assumptions on $Y$ (path connected, locally path connected, semilocally simply connected), it is known that semicoverings $f:X\to Y$ are the same as traditional covering maps. The proof is not entirely trivial because you need to verify path-homotopy lifting first. In fact, you can even do a little better. You can weaken "uniquely lifts all paths" to the following:
Theorem: Suppose $Y$ is path connected, locally path connected, and semilocally simply connected. A local homeomorphism $f:X\to Y$ is a covering map if and only if for every map $\beta:[0,1)\to X$ such that $f\circ \beta:[0,1)\to Y$ extends to a path $[0,1]\to Y$, $\beta$ also extends continuously to a path $[0,1]\to X$.
A proof follows from things in the original paper on semicoverings and the one I link to below. I don't know that a streamlined proof has been written anywhere.
I think the property given is easy to check in practice (it should feel like a kind of completeness condition on $X$ that depends on paths in $Y$). It is my understanding that some folks working in non-archimedian geometry have recently used these ideas. If you are unhappy with this property, then you'll probably need to be content with more restrictive topological conditions that imply it.
The necessecity of lifting paths from $Y$ to $X$ highlights that without extra compactness conditions, there must be some kind of relationship between the structure of $X$ and $Y$ to ensure that one actually has a covering map in hand. I have the feeling that a combination of basic topological conditions on $X$ or even the map $f$, which imply this lift-extension property is going to be quite restrictive and probably will put you close to assuming $f$ is proper anyway. For instance, if you assume $Y$ is locally nice in the usual way and that $X$ is sequentially compact, you can use this to conclude that $f$ is a covering map (see the paper When is a local homeomorphism a semicovering?). However, if $X$ is metrizable, then the sequentially compact condition is the same as $X$ being compact.
A: 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.
