Algebraic Varieties vs Smooth Manifolds There are many posts I have read on that subject which seem unclear for me. My main question (it might be silly) is: 

"Every non-singular algebraic variety over $\mathbb{C}$ is a smooth
  manifold."

(see: http://mathoverflow.net/questions/7439/algebraic-varieties-which-are-also-manifolds) 
How? For an algebraic variety we have the Zariski topology which is not even Hausdorff? How can they be diffeomorphic then?
 A: You are absolutely right that a complex variety with its Zariski topology is not a complex manifold, nor even a Hausdorff topological space (unless it has dimension zero).
However there is a completely canonical way of associating to a complex  algebraic variety $X$ a complex analytic variety $X^{an}$.
More precisely that association  Algvar $\to$ Anvar is a functor.
This functor has been studied in detail by Serre in a ground-breaking article published in 1956 and universally known by its amusing acronym GAGA.
A typical result in the article (Proposition 6, page12)  is that $X$ is complete iff $X^{an}$ is compact: a highly non-trival result relying on a theorem of Chow.   
In this set-up the result you are asking about can be stated as follows:
An algebraic complex variety $X$ is regular (=smooth) if and only if the associated analytic variety $X^{an}$ is a complex manifold.
Edit
Here is an English translation of GAGA.
A: The statements it that if the variety is endowed with the relative topology as a subset of $\Bbb C^n$ then it is a manifold.
