I have a solution to Exercise 4.2 in Hartshorne, which is essentially the same as the answer given here, but at the very end, I have a different argument which appears to remove the necessity of $X$ being reduced. Clearly, I have gone wrong somewhere, so I would like to know where the mistake is!
Consider the following diagram $$\require{AMScd} \begin{CD} X \times_{Y \times_S Y} Y @>{}>> Y\\ @V{d}VV @VV{\Delta_{Y/S}}V \\ X @>{(f\times_S g)}>> Y\times_S Y @>{p_2}>> Y\\ {} @V{p_1}VV @VVV \\ @. Y @>>> S \end{CD}$$ Once we have shown that $d$ is a surjective closed immersion, rather than arguing that $d$ is an isomorphism from the fact that $X$ is reduced, I would like to argue directly that $f$ and $g$ are the same morphism on $X$. The fact that $d$ is surjective implies that every point of $x$ has image in the diagonal. By this question, that means that for every $x \in X$, if $z = (f \times_S g)(x)$, we have that $p_1(z) = p_2(z)$, and further $(p_1)_z = (p_2)_z$, where these are the induced maps on stalks.
The first condition tells me that $f$ and $g$ are the same maps of topological spaces, and the second one says that $$\begin{align*} f_x & = ((f \times_S g) \circ p_2)_x \\ & = (p_2)_z \circ (f \times_S g)_x \\ & = (p_1)_z \circ (f \times_S g)_x \\ & = g_x. \end{align*}$$ Now, I know that one has to be careful concluding that two morphisms of sheaves are the same from their stalks, but in this case, it seems we have shown above that the induced maps on stalks are literally the same ring homomorphism at every point of $x$. Since the sections of a sheaf can be thought of as tuples of germs that are locally coherent (say via the construction of the associated sheaf in Hartshorne), and a morphism between the two should be determined by what it does to each stalk, shouldn't this imply that $f = g$?
Clearly, something has gone wrong, but I'm not entirely sure where. So where does the above argument fail?