This is a proof given in Gathmann's AG lecture notes for the regular functions on $U=A^2-\{(0,0)\}$.
Denote $D_f$ as the distinguished open set of function $f$.
It is clear that any regular function on $U$ has two representations as either $\frac{f_1}{x_1^n}$ or $\frac{f_2}{x_2^m}$. And two representation agrees on $D_{x_1}\cap D_{x_2}$. Then he concludes that $D_{x_1}\cap D_{x_2}\subset V(f_1x_2^m-f_2x_1^n)$ where the latter is closed and thus $Cl(D_{x_1}\cap D_{x_2})=A^2$. I do not see the last step.
My guess is that $A^2$ is irreducible. So denote $Cl(A)$ as closure of a set $A$. $Cl(D_{x_1}\cap D_{x_2}))\cup V(x_1x_2)=A^2$. So the closure must contain $A^2$ as $A^2$ is irreducible. Am I right on this part?
The other question is showing that if $U=X-V$ where $V$ is an irreducible codimension 2 subvariety of $X$, then $O_X(V)=A(X)$ where $A(X)$ is the coordinate ring of variety $X$. Why irreducible codimension 2 is required here?
Yes, your guess is correct. Since $\mathbb{A}^2$ is irreducible, any open set is dense for the reasons you mention.
The reason that codimension 2 is required is that the statement is false in codimension 1: take $\mathbb{A}^1\setminus{0}$, which has $k[x^{\pm1}]$ as it's regular functions.