Suppose I have a quasiprojective variety $X$ and a finite surjective map $$f: X \rightarrow Y$$ to a scheme $Y$. Is it true that $Y$ is quasiprojective as well? It seems like the answer could be no, but I don't know enough examples of nonprojective schemes.

This isn't true in general. For example, see Section 6 of Conducteur, Descente et Pincement by D. Ferrand. There Ferrand gives an example of a nonnormal proper variety $Y$ whose normalization is projective. If I recall correctly, many examples of proper nonprojective schemes have finite maps from projective ones. 


It is true if $Y$ is normal. Let me explain why. I will say that a variety has the ChevalleyKleiman (CK) property if every finite subset is contained in an affine open. By Corollary 48 of Kollar's "Quotients by finite equivalence relations" arXiv:0812.3608, if $f:X\to Y$ is finite and surjective, then if $X$ has the CK property, $Y$ has it too. Now, it is clear that a quasiprojective variety has the CK property, and a normal variety with the CK property is quasiprojective by Corollary 2 of "Quasiprojectivity of normal varieties" arXiv:1112.0975. Putting this together, we see that if $f:X\to Y$ is a finite surjective map of varieties with $X$ quasiprojective and $Y$ normal, then $Y$ is quasiprojective. 

