Does there exist a scheme not admitting a morphism locally of finite type to a quasi-compact scheme?

The reason I am asking this is that being locally of finite type and being quasi-compact are respectively most common local and global finiteness hypotheses in algebraic geometry and I think it is natural enough to wonder how they interact with each other.

A strictly stronger question has been asked and answered on MO (open immersions are, rather vacuously, locally of finite type).

Here is what I have tried. There exists a non-empty scheme $X$ having no closed points. Let $f:X\rightarrow Y$ be a morphism locally of finite type to a quasi-compact scheme. It is not too hard to show that there exists a closed point $p\in Y$. Consider the base change $f_p:X\times_Y \mathrm{Spec}\:k(p)\rightarrow X$ of $\mathrm{Spec}\:k(p)\rightarrow Y$. Being the base change of a closed immersion, $f_p$ is a closed immersion so it is enough to find a closed point in $X\times_Y \mathrm{Spec}\:k(p)$. The latter is a scheme locally of finite type over a field and as such has to have a closed point (first argue in the affine case by Zorn's lemma and then note that in this context, a point closed in an affine open is closed in the whole scheme).

The argument above is obviously incomplete because a morphism locally of finite type does not have to hit a closed point (consider e.g. $\mathrm{Spec}\:\mathbb{Q}_p\rightarrow \mathrm{Spec}\:\mathbb{Z}_p$). So it seems that one has to look for another obstruction.