I'll try to show that a filtered colimit of a diagram of subobjects is still a subobject, which I suppose should include your linearly ordered diagram as a special case.

A map $X\to I$ is a subobject (i.e., a monomorphism), if and only if the projection maps $X\times_I X\to X$ from the fiber product to either of the factors are isomorphisms.  If I have a family $X_i\to I$ of such monomorphisms indexed by a filtered category $C$, then $\mathrm{colim}_{i\in C} X_i \to \mathrm{colim}_{i\in C} I$ has to be monomorphism, since filtered colimits preserve finite limits, and because $\mathrm{colim}_{i\in C}I \approx I$ since $i\mapsto I$ is a constant diagram on the filtered category $C$. 

(Possibly, when I say "filtered", I really mean "cofiltered".  Can't figure out which is which.)

**Added.** I was thinking about a Grothendieck topos, which is cocomplete.  But it seems you (and Moerdijk and Mac Lane) want a proof which works in an elementary topos, and I don't know what to do in that case.

The key line in the proof seems to be "By completeness of $\mathrm{Sub}(I)$ ..."  Presumably, if $\mathrm{Sub}(I)$ is complete, we can form arbitrary "intersections" of subobjects; so $\mathrm{colim} X_i$ should be the intersection of all subobjects $Y$ of $I$ which contain all the $X_i$'s.

I don't know why $\mathrm{Sub}(I)$ should be complete in an elementary topos.  It is mentioned (for instance, on p. 491) that in a *cocomplete topos* $\mathrm{Sub}(I)$ is a Heyting algebra, so is complete.  So the proof should work in that case.