More precisely, we define $F:\mathbf{Sch}\to\mathbf{Sets}$ to be $F=\mathrm{forg}\circ\mathrm{sp}$, where $\mathbf{Sch}$ is the category of schemes, $\mathrm{sp}:\mathbf{Sch}\to\mathbf{Top}$ is the topological space functor, and $\mathrm{forg}:\mathbf{Top}\to\mathbf{Sets}$ is the forgetful functor. $F$ is said to be representable iff it is naturally isomorphic to $h^X = \mathrm{Mor}_\mathbf{Sch}(X,-)$.
I have no idea how to prove this.
Representable functors preserve all limits that exist in their domains, and in particular products. On the other hand, it is well known that the underlying space of the product of two schemes is not the product of the underlying of the factors — not as spaces and not even as sets.
All representable functors preserve products (in general, limts). So for a category like $\mathbf{Sch}$ in which the underlying set of the product is not the product of the underlying sets, the forgetful functor can't be representable.
If $F \simeq h^X$, then $$F(\operatorname{Spec}(\mathbb Z)) \simeq h^X(\operatorname{Spec}(\mathbb Z)) = \operatorname{Mor}(X, \operatorname{Spec}(\mathbb Z)).$$ But this is a contradiction because the right hand side is a singleton (since $\operatorname{Spec}(\mathbb Z)$ is final in the category of schemes), and the left hand side is not.