Hi all, I'm reading Mac Lane & Moerdijk's book "Sheaves in Geometry and Logic" and I don't understand a proof; sorry if this is the wrong place to ask, if there's somewhere better please let me know.

The proof in question is of Proposition VI.1.8 (page 276 in the paperback), which states:

Let $\mathcal{E}$ be a topos which is generated by subobjeccts of $1$, and moreover has the property that for each object $E$, $\mathrm{Sub}(E)$ is a complete Boolean algebra. Then $\mathcal{E}$ satisfies the axiom of choice.

Here the axiom of choice is the statement that any epimorphism $p: X \to I$ has a section $s: I \to X$, i.e. $ps = 1_I$. The proof starts as follows:

Let $p: X \to I$ be an epimorphism in $\mathcal{E}$. By completeness of $\mathrm{Sub}(I)$, we can apply Zorn's lemma and find a maximal subobject $m: M \to I$ such that $p$ has a section $s: M \to X$, i.e., $ps = m$.

I don't see how to show that Zorn's lemma applies here. Presumably I need to show that, given a linearly ordered set of subobjects $m_i: M_i \to I$ of $I$, $m_i = m_j k_{ij}$ for some $k_{ij}: M_i \to M_j$ whenever $m_i \leq m_j$, together with a set of sections $s_i: M_i \to X$ such that $p s_i = m_i$, $s_j k_{ij} = s_k$, there exists a section $s: M \to X$ such that $ps= m$, where $m: M \to I$ is the least upper bound in $\mathrm{Sub}(I)$ of the $m_i$'s, and such that the restriction of $s$ to each $M_i$ is $s_i$. It seems plausible that $m: M \to I$ will turn out to be a colimit in $\mathcal{E}/I$ of the collection $m_i: M_i \to I$, from which the required result would follow easily, but I don't know how to prove this. Can anybody help?