Problem in proving that $\mathbb{A}^2$ is not homeomorphic to $\mathbb{P}^2$ Let $k$ be an algebraically closed field. All spaces are equipped with the usual Zariski topologies. 

All the proofs of this fact that I've seen rely on the fact that two lines in $\mathbb{P}^2$ intersect but this doesn't necessarily hold in $\mathbb{A}^2$. I'm stuck on proving that this property is a "Zariski-topology invariant" (i.e preserved by homeomorphism). All the proofs use this fact without proving it, so I assume it is trivial, but I do not know how to prove it.

Does someone have a hint on how to prove it?

It would be enough for me to prove that lines in $\mathbb{A}^2$ are sent to projective lines to complete the proof, or that the image of an algebraic curve is an algebraic projective curve. But I cannot prove any of these. Any help?
There exist two disjoint irreducible  closed subsets  both containing more than one point   in $\mathbb A^2_k$ but not in $\mathbb P^2_k$ (Bézout).
Well, there are two things.


Homeomorphism preserves compactness (can you see why? homeomorphisms being one-to-one preserves unions, inclusions and opennes).
$\mathbb{P}^2$ is compact, while $\mathbb{A}^2$ is not.