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?
[Math] When is a compact topological 4-manifold a CW complex
4-manifoldsat.algebraic-topologycw-complexesopen-problemstopological-manifolds
Best Answer
Hatcher, Algebraic Topology, Corolary A.9
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?