If $(X,\mathscr O_X)$ is a scheme and $U$ an open subset of $X$, how does it follow that $(U,\mathscr O_X{_{|U}})$ is a scheme? I found this as a remark in Bosch's book, 'Algebraic Geometry and Commutative Algebra' just after the definition of a scheme.
I don't get this even when the scheme is affine(except when the open set is basic).
Pick an affine cover $\left\{ {{\mathcal{U}_i}} \right\}$ of $X$ and a cover by distinguished open sets $\left\{ {\operatorname{D} \left( {{f_{ij}}} \right)} \right\}$ for each $\mathcal{U}_i$.
Then for any open subset $\mathcal{U}$ of X, $\mathcal{U} \cap {\mathcal{U}_i}$ is an open subset of the affine $\mathcal{U}_i$. Thus $\mathcal{U} \cap {\mathcal{U}_i}$ is covered by distinguished affines $\operatorname{D} \left( {{f_{ij}}} \right)$ for some j's.
And so $\mathcal{U}$ has an affine cover by $\operatorname{D} \left( {{f_{ij}}} \right)$ for some i,j's.