I was reading article 4.4.3 from Ravi Vakil's Algebraic Geometry book: http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf, which got me rather confused. I am aware that many similar questions have been asked previously, but none of them have been able to satisfactorily clarify if what I am thinking is either wrong or correct. I would be greatly obliged if someone could point out the flaws in my thinking.
First, I understand how $\mathcal{O}_{\mathbb{A}^2}\mid_U(U)$ is precisely $k[x,y]$. Now, because of this, if $U$ is to be isomorphic to some $(\operatorname{Spec}(A),\mathcal{O}_{\operatorname{Spec}(A)})$, we would need $A=k[x',y']$ (this is the same thing as $k[x,y]$, but I'm using slightly different notation as I don't immediately see why the isomorphism we'll talk about is the canonical one). Now, we have an isomorphism of rings: $$k[x,y]\to k[x',y']$$
This induces a homeomorphism $$\mathcal{O}_{\mathbb{A}^2}\mid_U(U) = \operatorname{Spec}(k[x,y]) \to \operatorname{Spec} (k[x',y'])$$ We consider the prime ideal $\mathfrak{p}=(x,y)\subset k[x,y]$. Suppose the ring isomorphism maps $\mathfrak{p}$ to $\mathfrak{p}'\subset k[x',y']$. This means that the induced homeomorphism maps $V(\mathfrak{p})\subset U$ to $V(\mathfrak{p}')\subset \operatorname{Spec}(k[x',y'])$. But this gives us a contradiction, for the former is empty (as discussed in Vakil) while the latter clearly isn't, and we are done.