We say $X$ is a quasicompact scheme (qc) if every open cover has a finite subcover. We also say that a morphism $f:X\rightarrow Y$ is quasicompact if $f^{-1}(U)$ is qc for every affine open set $U$ of $Y$.
Why is $X$ being qc equivalent to the morphism $X\rightarrow\operatorname{Spec}\mathbb{Z}$ being qc? Clearly, if $f$ is quasicompact, then $f^{-1}(\operatorname{Spec}\mathbb{Z})=X$ is quasicompact. But the opposite direction is what gives me trouble.
I know that $\operatorname{Spec}\mathbb{Z}$ is the final object in the category of schemes, but I'm not sure where to go from there.
As @KReiser pointed out in the comment if $f:X\rightarrow \operatorname{Spec}\mathbb Z$ is quasi-compact, then $f^{-1}(\operatorname {Spec}\mathbb Z)=X$ is a quasi-compact topological space.
Conversely assume $X$ is a quasi-compact scheme. Any open affine subset of $\operatorname{Spec}\mathbb Z$ is a finite union of distinguished open affines $D(n)$ for $n\in \mathbb Z$. So if $X=\bigcup_{i\in \mathscr F}X_i$ is a finite union of open affines, then $f^{-1}(D(n))=\bigcup_{i\in \mathscr F}{X_i}_{f^\#(n)}$ which is a finite union of open affines and hence quasi-compact. (${X_i}_{f^\#(n)}$ is a distinguished open set subset of the affine scheme $X_i$ and hence affine) Thus $f^{-1}(U)$ is quasi-compact for every open affine $U\subset \operatorname{Spec}\mathbb Z$. So $f$ is quasi-compact.