What is a non-logical concrete example of a choiceless elementary topos? What is a non-logical concrete example of a choiceless elementary topos?
The elementary topoi made from choiceless models of ZF, or stuff like the realizability topos, are (I believe) choiceless, but is there a concrete example from other branches of mathematics?
 A: Assuming you mean choice in the "every epimorphism has a section" sense, counterexamples among categories of presheaves are plentiful.
For a very quick and silly example, there's the category $\mathbf{Set}^G$ for a non-trivial group $G$. In this case, the sole representable functor in the category is the transitive action of the group on itself, so it has no fixed points; so there is no section of the unique map from this object to the terminal object of $\mathbf{Set}^G$.
Another, similar example is the category $\mathbf{Set}^{\bullet\rightrightarrows\bullet}$ of directed multigraphs. Consider the graph with two vertices, and one arrow in each direction between them; then the unique map from this to the terminal object also doesn't have a section.
More generally, toposes satisfying choice have a classical internal logic, so choosing any topos with a strictly intuitionistic internal logic will involve some failure of choice. You can hardly throw a stone without hitting an example of a topos where choice fails.
A: Consider the topos corresponding to the topological space $\mathbb{C}$, with the object $\mathfrak{S}$ being the sheaf of analytic functions.  Then the differentiation operator $D : \mathfrak{S} \to \mathfrak{S}$ is an epimorphism (essentially because $\mathbb{C}$ is locally simply connected) - so the interpretation of the formula $\forall f \in \mathfrak{S}, \exists F \in \mathfrak{S}, D(F) = f$ is $\top$.  However, because any nonempty open subset of $\mathbb{C}$ contains a punctured open disc, $D$ does not have a right inverse when restricted to any nonempty open subset; therefore, the interpretation of the formula $\exists I \in \mathfrak{S}^{\mathfrak{S}}, D \circ I = \operatorname{id}_{\mathfrak{S}}$ is $\bot$.
