It seem to me that the problem that Makkai has in mind is that the existence of non-trivial choice objects is in conflict with non-booleaness.

The core of the arguement, is the following lemma, which essentially follows from Diacunescu's proof that $AC \Rightarrow LEM$:

**Lemma:** Let $A$ be a choice object in a topos, then internally:

$$ \forall x,y \in A, \forall U \in \Omega, (U \Rightarrow (x=y)) \cup U $$

**Remark:** that is basically a rephrasing of lemma D4.5.11 of P.T.Johnstone sketches of an elephant.

**Proof:** Let $A$ be a choice object and $x, y \in A$. Let $U$ be any proposition, consider the set $1 \coprod_U 1$ and the relation that send the first component to $x$ and the second to $y$ (and to both for element that are in both component). It is an entire relation to $A$, so there is function $1 \coprod_U 1 \rightarrow A$ included in that relation, in particular which takes values in $\{x,y\}$. I'm calling $a$ and $b$ the "two" elements of $1 \coprod_U 1$, I have four case to deal with regarding the values of $a$ and $b$ by this functions, but in each of them you can either prove $U$ or $U \Rightarrow (x,y)$:

Geometrically speaking, this is a pretty rough restriction on what can choice objects be ! it means that if $A$ is a choice object in a topos $\mathcal{T}$, and $a,b$ are two section of $A$ on $X$, then the closed complement of $(x=y)$ in $\mathcal{T} / X$ is Boolean. So either $A$ is very close to be subterminal, or the slices of $\mathcal{T}$ have some large boolean closed subtoposes. (this is the conflict I was referring to at the begining)

But one can do better:

**Proposition:** Let $\mathcal{T}$ be a Grothendieck topos which:

then $\mathcal{T}$ is degenerate. For example, $Sh([0,1])$ do not satisfies SVC.

Note that a nowhere boolean topos is a topos where LEM is internally false, in the sense that the interpretation of "$\forall U, U \cup \neg U$" in the internal logic is $\bot$ (the initial object).

We start with another lemma:

**Lemma:** In a nowhere boolean topos, if $A$ is a choice object and $D$ is decidable object (i.e. internally $\forall x,y \in D, x=y \cup x \neq y$) then one can internally show that:

$$ \texttt{Any partial map $A \rightarrow D$ is constant} $$

Indeed, consider internally a partial map $f:A \rightarrow D$, for each $a , a'$ in the domain of definition of $f$, either $f(a)=f(a')$ or $f(a) \neq f(a')$, but $f(a) \neq f(a')$ implies $a \neq a'$, which internally implies LEM by the previous lemma, which is false as $\mathcal{T}$ is nowhere boolean. Hence the result.

One can now prove the proposition: If one assumes further that $\mathcal{T}$ satisfies SVC, it means that every decidable object of $\mathcal{T}$ is a subquotient of the object $S$ (as the lemma above implies that the partial map $S \times A \rightarrow D$ is constant in the $A$-direction). But if $\mathcal{T}$ is non degenerate, for $\kappa$ large enough (larger than the size of the site and the size of $S$), the object $p^* \kappa$ is locally decidable and of size $\kappa$, so it cannot be a subquotient of $S$.

**Edit:** *My apologies for the needlessly non mathematical and complicated first answer. I think I found the theorem I was after.*

Grothendieck toposeswhere it fails. If you don't know what a Grothendieck topos is, you can think of it as a category-theoretic version of the forcing model over a set-sized site (mathoverflow.net/q/13480/49). Note that when we allow arbitrary sites, the result in general models only IZF rather than ZFC. $\endgroup$ – Mike Shulman Aug 28 '19 at 0:118more comments