If $X, Y$ are topological spaces, with $Y$ being a k-space, and $f : X \to Y$ is a proper covering map, is $X$ necessarily a k-space? A k-space is a compactly generated Hausdorff topological space. (I used the terminology "k-space" in the question, in order keep the question within the limit of 150 characters.) 
Note that under the stated hypotheses in the question, $X$ is automatically Hausdorff and $f$ is automatically a closed map.
Thus, in more basic terms, the question is: If $X$ and $Y$ are topological spaces, with $Y$ being compactly generated and Hausdorff, and $f : X \to Y$ is a proper covering map, then is $X$ necessarily compactly generated? 
If the answer is negative, is there a positive answer if $X$ is also assumed to be connected and locally path-connected (both being common assumptions on covering spaces)?
A particular case, in which the answer is positive, is when $Y$ is (Hausdorff and) locally compact. In this case, even fewer hypotheses suffice; for it is well known that if $X$ and $Y$ are Hausdorff, with $Y$ being locally compact, and $f : X \to Y$ is a continuous, proper, surjection, then $X$ is also locally compact. (The converse is also true). Does this result also hold if we replace the hypothesis, "locally compact", with "compactly generated"?
 A: In the question stated the map $f$ has two properties: being proper and being a covering map.
In fact, each of these properties by itself is enough to deduce that $X$ is a k-space.
The comment by @erz explains that being a covering map is enough:
a local $k$-space is a k-space and a covering space of a k-space is a local k-space.
I am writing this to document the complementary fact that also the properness of $f$ alone is enough to deduce that $X$ is a k-space.
Claim:
A (weakly) Hausdorff topological space admitting a continuous proper map to a (weakly) Hausdorff compactly generated topological space is compactly generated.
The proof of the claim is based on the following easy lemma which I leave as an exercise (but you can have a peek at Lemma 2.13 here).
Lemma: If $g:A\to Y$ is closed and proper and $x_\alpha$ is a net in $A$ such that $g(x_\alpha)$ converges in $Y$ then $x_\alpha$ has a converging subnet.
Proof of the claim: Assume $f:X\to Y$ is a continuous proper map, $X$ is Hausdorff and $Y$ is Hausdorff compactly generated. Fix $A\subset X$. 
As $X$ is Hausdorff it is clear that for every compact subset $K\subset X$, $A\cap K$ is closed. We will prove the converse. Assume by contradiction that $A$ is not closed and for every compact subset $K\subset X$, $A\cap K$ is closed. Let $x_\alpha$ be a net in $A$ converging to a point not in $A$.
Note that $g=f|_A:A\to Y$ is a proper map: for $K\subset Y$, $f^{-1}(K)$ is compact, thus $g^{-1}(K)=f^{-1}(K)\cap A$ is compact too, as it is closed and contained in $f^{-1}(K)$.
As $Y$ is a Hausdorff k-space, we get that $g$ is also closed.
As $x_\alpha$ has no conveging subnet in $A$, but $g(x_\alpha)$ converges in $Y$
we get a contradiction to the lemma.
