You can suppose $Y$ is affine. Cover $X$ by affine open subsets {$U_i$}$_i$. As $Y$ is quasi-compact, a finite number of the $f(U_i)$ cover $Y$. The union of the correponding (finitely many) $U_i$ is the $U$ you want.

[Edit] (add some details).

Replacing $Y$ with an affine open neighborhood $V$ of $f(x)$ (and $X$ with $f^{-1}(V)$), one can suppose that $Y$ is affine. Cover $X$ by affine open subsets {$U_i$}$_i$. As $Y$ is quasi-compact, a finite number of the $f(U_i)$ cover $Y$. If necessarily, we can add one more $U_i$ so $x$ belong to one of these $U_i$'s. The union $U$ of these (finitely many) $U_i$ is quasi-compact, and we have $f(U)=Y$, $x\in U$. The morphism $f|_U : U\to Y$ is a morphism from a quasi-compact scheme to an affine scheme, so it is quasi-compact because for any affine open subset $V$ of $Y$, $(f|_U)^{-1}(V)\cap U_i= V\times_Y U_i$ is affine.