Regular functions on plane minus origin Let $k$ be algebraically closed and consider $X = \mathbb{A}^2 \setminus O$. How does one show that $\mathcal{O}(X) \cong k[x, y]$?
I am more or less aware of how to do this from a sheaf-theoretic point of view: the distinguished opens are $D(x)$ and $D(y)$, and the regular functions on these are simply elements of $k[x, y]_x$ and $k[x, y]_y$, so a regular function is a function which agrees on the overlap, and thus has no power of $x$ or $y$ in the denominator, i.e. is an element of $k[x, y]$.
Can this proof be translated to the classical point of view? In particular, how does one assert that the regular functions on a "distinguished open set $D(f)$" are simply the polynomials localized at $f$ -- the rest will be the same after this. 
 A: I'll answer your main question:

How does one assert that the regular functions on a "distinguished open set $D(f)$" are simply the polynomials localized at $f$?

As you said, everything else follows from this.
Before we can begin, we must agree on a "classical" definition of regular functions on open sets. Assuming that $X$ is an irreducible affine variety $V(I)$, my preferred definition is as follows:


*

*We will say that a tuple $(U, f / g)$ is "valid" iff $U$ is a non-empty open subset of $X$, and $f$ and $g$ are elements of $A := k[x_1, \dots, x_n]/I$ such that $g(x) \neq 0$ for all $x \in U$.

*We now define an equivalence relation on the set of valid tuples, where $$(U, f / g)\sim (V, f' / g') \ \ \ \iff \ \ \ \tfrac{f(x) }{ g(x)} = \tfrac{f'(x)}{ g'(x)} {\rm \ for \ all \ } x \in U \cap V.$$
For this to work, it is vital that $X$ is irreducible! Irreducibility ensures that:

If $\frac{f(x)}{ g(x)} = \frac{f'(x) }{g'(x)}$ holds for all $x$ in some non-empty open subset of $U \cap V$, then $\frac{f(x)}{ g(x)} = \frac{f'(x)}{ g'(x)}$ holds for all $x \in U \cap V$.

$ \ \ \ \ \ \ \ \ $This "little lemma" guarantees that the equivalence relation really is transitive.


*

*We say that a rational function on $X$ is an equivalence class, $[(U, f / g)]$.

*We can define an field structure on our set of rational functions in the obvious way. For example $$[(U, f / g)] + [(V, f'/g')] = [(U \cap V, (fg' + f'g)/gg')]$$ Our "little lemma" ensures that this definition is independent of the chosen representatives of the equivalence class. 

*Now observe that our field of rational functions is isomorphic to the fraction field ${\rm Frac}(A)$. The isomorphism is simply
$$ \phi :  [(U, f/g)] \mapsto \tfrac{f}{g} \in {\rm Frac}(A)$$
The fact that $\phi$ is independent of the choice of representative for the equivalence class follows from our "little lemma". So too does the injectivity of $\phi$.

*We say that a rational function is regular at a point $x \in X$ iff there exists a representative of the equivalence class of the form $(U, f/g)$, where $x \in U$.

*Clearly, a rational function $[(U,  f / g)]$ is regular at $x$ if and only if
$$ \phi \left( [(U, f / g)] \right) \in A_{\mathfrak m_x} \subset {\rm Frac}(A)$$
where $A_{\mathfrak m_x}$ is localisation of the ring $A$ at the maximal ideal associated to the point $x$.
So to wrap up: what does it mean for a rational function to be regular at all points in $D(f)$? It means the image of our rational function under the isomorphism $\phi$ is contained in the intersection
$$ \bigcap_{x \in D(f)} A_{\mathfrak m_x} \subset {\rm Frac}(A).$$
This intersection is equal to
$$ \bigcap_{\mathfrak m {\rm \ maximal \ } \\ \ \ \ \ \ \ \subset \  A_f} (A_f)_{\mathfrak m} \subset {\rm Frac}(A_f) \cong {\rm Frac}(A).$$
And by a standard result in commutative algebra, this intersection is precisely the ring $A_f$. So it is in this sense the ring of regular functions on $D(f)$ is identified with $A_f$.
Returning to the problem about $\mathbb A^2 \ \backslash \ \{(0, 0\}$, we are asked to find all rational functions that are regular at all points in $D(x)$ and are also regular at all points in $D(y)$. So we must find the intersection of the rings $k[x,y]_x$ and $k[x,y]_y$ within the fraction field ${\rm Frac}(k[x,y])$, and this intersection is precisely $k[x,y]$.
