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Freedman's $E_8$-manifold is nontriangulable, as proved on page (xvi) of the Akbulut-McCarthy 1990 Princeton Mathematical Notes "Casson's invariant for oriented homology 3-spheres". Kirby showed that a compact 4-manifold has a handlebody structure if and only if it is smoothable: 1 and 2. When is a compact topological 4-manifold a CW complex?

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Andrew Ranicki
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    I take it this is an open problem? Small note, but your question was effectively asked in another form, here: http://mathoverflow.net/questions/36838/are-non-pl-manifolds-cw-complexes I usually think of CW-complexes as being a tool for describing homotopy-types rather than homeomorphism types, so my answer was to a weaker question than the one asked. – Ryan Budney Aug 22 '11 at 19:09
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    I have good reason to believe that it is an open question! Apologies - I hadn't seen the earlier posting http://mathoverflow.net/questions/36838/are-non-pl-manifolds-cw-complexes – Andrew Ranicki Aug 22 '11 at 21:56

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Hatcher, Algebraic Topology, Corolary A.9

Every compact manifold, with or without boundary, is an E[uclidean]N[eighborhood]R[etract].

See also the following three numbered conclusions.

Or, in simpler terms, CW is a lot more flexible than, say, PL.

By "compact manifold", I expect you mean, in particular, an object $X$ with a finite very good cover --- where, in case I need to be so careful, by "very" good cover, I mean that the boundary of every intersection of the Čech complex is a sphere. Without Loss Of Generality, every compact manifold has such a cover.

So fix one finite good cover, and refine it to a finite good cover by closed sets. Construe the Čech complex of this good finite closed cover as a simplicial-set-over $X$. Its realization has the homotopy type of both $X$ (it is fiberwise finite contractible) and a CW complex: roughly, each open patch of the Čech complex is a cone over a sphere, and there are finitely many, so we are iteratively gluing in a finite family of products $\triangle^n \times C(\mathbb{S}^k)$. This doesn't literally produce a CW complex, in that the attaching of cells isn't in cellular order; however it is a finite process, amenable to iterative correction, and the Reader can supply the details as needed.

Since you mention Handlebodies I wonder: are you actually trying to get at which $4$-manifolds are homeomorphically 4-dimensional CW-complexes?

Jesse C. McKeown
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    Yes, the point is to determine when a 4-manifold is homeomorphic to a CW complex. See also Question 1.3 here – Dario Aug 11 '15 at 07:01
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    A compact 4-manifold is a compact topological space which is locally homeomorphic to R^4. Borsuk proved in the 1930's that compact topological manifolds (in all dimensions) are ENR's. But without differentiable structure, it is by no means clear that a topological manifold has a good cover. And yes, I do mean to get at compact 4-manifolds which are homeomorphic to (and so are) 4-dimensional CW complexes. – Andrew Ranicki Aug 12 '15 at 06:23