$\newcommand{\P}{\mathbb{P}_k}$ $\newcommand{\A}{\mathbb{A}_k}$ I am trying to find another way of showing that $\P^2$ is not homeomorphic to $\A^2$... namely I am trying to show that the cohomologies with respect to the constant sheaf is zero.
The usual cover $U_0,U_1,U_2$ is not a good cover... the cohomology is concentrated in degree zero with respect to this cover. ( If you look in the $\P^1$ when $k=\mathbb{C}$ case you see that $U_0 \cap U_1$ has nontrivial cohomology with respect to the constant sheaf. Hence this is why the cohomology ends up being zero).
Question: What is a good cover for $\P^2$?
Any constant sheaf on an irreducible topological space is flasque. Specifically, if $\mathcal{F}$ is the sheaf of locally constant maps to a set $A$ on a space $X$, then $\mathcal{F}(U)$ consists of just the constant maps if $U$ is a connected open set. But if $X$ is irreducible, every open subset is connected, and so $\mathcal{F}(U)$ is just the constant maps for all $U$. It follows that all the restriction maps $\mathcal{F}(U)\to\mathcal{F}(V)$ are bijections (unless $V$ is empty), and $\mathcal{F}$ is flasque.
In particular, if $A$ is an abelian group, the cohomlogy of $\mathcal{F}$ will be trivial in positive degrees. This applies both to $X=\mathbb{A}^2_k$ and $X=\mathbb{P}^2_k$, and so cannot be used to distinguish them topologically.