Let $X$ be a quasi projective variety. The function field of $X$, $k(X)$, is defined as $k(X\cap U)$ where $U$ is any affine open subset of $X$ and $k(U)$ is the function field of the affine variety, which is well defined.
By question is, how to show that $k(X)$ is independent of the choice of $U$?
By definition, a rational function on an irreducible algebraic variety $X$ over a field $k$ is an equivalence class of morphisms $U\to k$, where $U$ is an open subset of $X$. We say that two morphisms $U\to k$ and $U'\to k$ are the same if they coincide on the intersection.
With this definition, you see that the field of rational functions is the same if your restrict to an open subset. You can thus take an open affine subset, and look at it on this.
What is good with affine algebraic varieties, is that they are the Spectrum of a ring $A$ and that regular functions correspond to elements of $A$. The variety is irreducible if and only if the ring $A$ is integral, and in this case you can check with the above definition that rational functions are just elements of the field of fractions of $A$. Hence, looking at affine open subsets is interesting as you can compute easily the field of rational functions.
Moreover, taking the above definition, you directly see that the choice is independent of the open subset, and thus of the affine open subset.