Apparently, there is an [abstract nonsense argument](http://mathoverflow.net/a/159995/11640) that shows $\mathbf{Sch}$ is concretisable. Here is a hands-on proof. 

We define $U_0 : \mathbf{Sch} \to \mathbf{Set}$ to be the functor that sends a scheme to the set of points of the underlying topological space and we define $U_1 : \mathbf{Sch} \to \mathbf{Set}^\mathrm{op}$ to be the functor that sends a scheme to the disjoint union of the stalks of its structure sheaf. (This makes sense because the stalk of $f^{-1} \mathscr{O}_Y$ at $x$ is the stalk of $\mathscr{O}_Y$ at $f (x)$.) Clearly, $(U_0, U_1) : \mathbf{Sch} \to \mathbf{Set} \times \mathbf{Set}^\mathrm{op}$ is faithful, and the contravariant powerset functor $\mathscr{P} : \mathbf{Set}^\mathrm{op} \to \mathbf{Set}$ is also faithful, so the functor $X \mapsto U_0 X \amalg \mathscr{P} (U_1 X)$ is a faithful functor $\mathbf{Sch} \to \mathbf{Set}$.