Let $k$ be either the field $\Bbb C$ of complex numbers or the field $\Bbb R$ of real numbers.
Let $X$ be an algebraic variety over $k$, say, quasi-projective and smooth (but not necessarily projective).
We consider the set of $k$-points $X(k)$ with the usual topology.



> 
> **Question.** Is $X(k)$ homotopically equivalent to a CW complex?
> 
> 
> 



Is a complex or real algebraic variety homotopically equivalent to a CW complex?
This is same answer I left at the question linked above in the comments. The one benefit it offers is that it directly addresses the class of varieties in question.


In *[Triangulation of Locally Semi-Algebraic Spaces.](https://deepblue.lib.umich.edu/handle/2027.42/63851)* by K.R. Hofmann, necessary and sufficient conditions are given for a *locally* semi-algebraic space to be homeomorphic to a simplicial complex, with a corollary that any abstract algebraic variety over $\mathbb{R}$ or $\mathbb{C}$ admits a triangulation.


Here is an excerpt from the abstract:



> 
> "We give necessary and sufficient conditions for a locally
> semi-algebraic space to be homeomorphic to a simplicial complex. Our
> proof does not require the space to be embedded anywhere, and it
> requires neither compactness nor projectivity of the space. A
> corollary is that every real or complex algebraic variety is
> triangulable, a result which does not seem to be available in the
> literature when the variety is neither projective nor real and
> compact."
> 
> 
> 


