Yes, you can do it over any field.

First, it is enough to show $\mathcal O(Y) = k[x,y]$ ($Y=A^2-0$). If that is true and $Y$ is affine, then the embedding $Y \to A^2$ must correspond to some $k$-algebra map $k[x,y] \to k[x,y]$, which is absurd.

The key point now, as in Guillermo's post, is to show that $R_{(x)} \cap R_{(y)}= R$ ($R=k[x,y]$). It will follow from the

Fact: $(x^m, y^n)$ form a regular sequence on $R$ for all $m,n>0$

Indeed, if $f/x^m =g/y^n$, then $fy^n=0$ modulo $x^m$, so $f=hx^m$ and we are done.

The above Fact is elementary. For example you can induct on $m$. Clearly $m=1$ is OK. Now if $m>1$, use the short exact sequence:

$$0 \to R/{(x^{m-1})} \to R/(x^m) \to R/(x) \to 0$$

and Snake Lemma to conclude that $y^n$ is regular on the middle term as well.

Note that I used $x,y$ abstractly and all you need is that the elements $(x,y)$ form a regular sequence on a commutative Noetherian ring $R$ to start with. Then the proof shows that $\mathcal O(Y) =R$ if $Y=\text{Spec}(R) - V(x,y)$. In fact, more general results are true for ideals of depth at least $2$. If you are interested, it will be a good motivation to learn about depth and regular sequences.