We can easily see that the function field of $\mathbb{A}^2_k-(0,0)$ is still $k(x,y)$. So the ring of functions is of the form $f/g$ where $f$ and $g$ are polynomials. But any polynomial in 2 variables will vanish at a codim 1 sub-variety, i.e. cannot vanish at exactly 1 point. This is the Krull dimension theorem. But if you think this is too much, from the fact that $k$ is algebraically closed, you can see that $g$ must vanish at more than 1 point: for each $x$ you can solve for $y$. Thus the ring of functions on $\mathbb{A}^2_k-(0,0)$ is $k[x,y]$. Thus, if it's affine, it must be isomorphic to $\mathbb{A}^2_k$ through the identity map. But it's not. So we are done.
Another way that uses Cohomology is the follows: using $\check{\mathrm{C}}\mathrm{ech}$ cohomology, we can show that $H^1(\mathbb{A}_k^2-\{0\}, \mathcal{O}_X)$ is infinite dimensional. But if our space is in fact affine, then this must vanish, due to Serre's criterion for affineness.
We can easily see that the function field of $\mathbb{A}^2_k-(0,0)$ is still $k(x,y)$. So the ring of functions is of the form $f/g$ where $f$ and $g$ are polynomials. But any polynomial in 2 variables will vanish at a codim 1 sub-variety, i.e. cannot vanish at exactly 1 point. This is the Krull dimension theorem. But if you think this is too much, from the fact that $k$ is algebraically closed, you can see that $g$ must vanish at more than 1 point: for each $x$ you can solve for $y$. Thus the ring of functions on $\mathbb{A}^2_k-(0,0)$ is $k[x,y]$. Thus, if it's affine, it must be isomorphic to $\mathbb{A}^2_k$ through the identity map. But it's not. So we are done.