I know that this obviously isn't true, but I can't find a gap in my "proof" and it's really bothering me. Here, "variety" means reduced, separated, finite type scheme over some field $k$.
If $X$ is some variety over $k$, then the morphism $X \to \text{spec} \, k$ is finite type, thus quasicompact. Thus the preimage of open affine $\text{spec} \, k$ is quasicompact, yet this preimage is $X$, which would imply that any variety is quasicompact, which is certainly not true.
Can someone please help me figure out what is wrong with my reasoning?