Why is this map almost an étale covering? Let $V$ and $W$ be absolutely irreducible quasi-projective varieties over $K$ with char $K=0$ with dim $V=$ dim $W$ and let $f: W \rightarrow V $ be a dominant morphism.
Why is it possible to make $f$ into an étale covering by restricting $V$ ?  
I know that this is probably a very basic and silly question but I have only started reading about algebraic geometry. 
 A: Generically, the map is etale. That is, there exists a nonempty open set $U \subset V$ such that $f^{-1}(U) \to U$ is finite and etale. Indeed, we can assume $f$ is flat at the outset by the generic flatness theorem. Then this follows because if $\xi$ is the generic point of $V$, $f^{-1}(\xi) \to \xi$ is a finite etale morphism (it's the extension of function fields, which is finite and separable as we are in characteristic zero).
Alternatively, the sheaf $\Omega_{W/V}$ is zero at the generic point of $W$ for the same reasons above, so its support is a proper closed subset $Z \subset W$. If we take $U = V- \overline{f(Z)}$, then the map $f^{-1}(U) \to U$ is unramified (and as above we can shrink further, if necessary, to get flatness).
Finally, to get finiteness, again we observe that $f^{-1}(\xi) \to \xi$ is finite, and since $\xi$ is the inverse limit over all nonempty open sets $U \subset V$, it follows that $f^{-1}(U) \to U$ is finite for some $U$. (This is a special case of the "noetherian descent" argument expounded at length in EGA IV-8.)
