Subfunctor of a hom functor in a Boolean topos

Question

There is a wrong argument below. The conclusion is not correct but I can't figure out why. Could anyone tell me why it is not correct?

Let X be any space and suppose the category of all sheaves $Sh(X)$ is Boolean. Let $1=Hom(-,X)$, then the subobject of $1$ in $Sh(X)$ can only be itself or $0$ as every subobject of $1$ gives a sieve, either the subobject or its complement contains $id_X$ hence contains the whole $Hom$ functor. If so, by the fact that there is an isomorphism $\mathcal{O}(X) \cong Sub_{Sh(X)}(1)$, there are only two elements in $\mathcal{O}(X)$ but this conclusion is not true.

Answer 1

either the subobject or its complement contains $id_X$

This is false. For instance, if $X=\{0,1\}$, then the subobject $\operatorname{Hom}(-,\{0\})$ has complement $\operatorname{Hom}(-,\{1\})$, but neither contains $id_X$. What's going on is that the join of these subobjects can contain $id_X$ even though neither one does on its own, by the gluing axiom for sheaves.