Covering image of a connected CW-complex need not be a CW-complex.

Problem: Let $$X$$ be a connected CW-complex, and $$Y$$ be a connected topological space. Suppose $$p: X\to Y$$ is a covering map. Does there exist a CW-structure on $$Y$$? More generally, is $$Y$$ homotopically equivalent to a CW-complex?

Note that the other direction is clear: Any covering of a connected CW-complex can always be given a CW-structure, lifting the characteristics map of cells of base space such that the covering map is a cellular map.

I believe that the answer to the above problem is no, but I have no counterexample.

$$\bullet$$ Notice that $$Y$$ is locally path-connected as the covering map is a local homeomorphism, hence $$Y$$ is path-connected also. So, we can not consider spaces $$\{0\}\cup\left\{\frac{1}{n}:n\in\Bbb N\right\}$$ or Topologist Sine Curve as $$Y$$. Notice that both $$\{0\}\cup\left\{\frac{1}{n}:n\in\Bbb N\right\}$$ or Topologist Sine curve are not homotopically equivalent to a CW-complex.

$$\bullet$$ Similarly, we can not consider the Hawaiian Earring(this is not semi-locally simply connected) as $$Y$$: The connected CW-complex $$X$$ has the universal cover so that $$X$$ is semi-locally simply connected, but the property "semi-locally simply connected" is preserved under a local homeomorphism.

So, I am run out of examples. Any help will be appreciated. Thanks in advance.

As far as I know this is open, but assuming the conjecture is true the answer to your question is that a CW complex may cover a non CW complex. Let $$M$$ be a nonsmoothable 4-manifold with infinite fundamental group, these are known to exist. Let $$\widetilde{M}$$ denote its universal cover. $$\widetilde{M}$$ is noncompact, and it is known that noncompact 4-manifolds are triangulated. Hence, $$\widetilde{M}$$ has a CW structure and it covers $$M$$ which is not a CW complex, if Ranicki's conjecture is true.