I've been revisiting the chapter on spectral sequences in Weibel's An Introduction to Homological Algebra and trying to pay attention to how everything works out in arbitrary abelian categories as opposed to the more concrete categories of modules. The first complication arises in 5.2.8, when the limiting terms are introduced. I will use homological indexing.
Say we have a spectral sequence $E$ starting at the $a$-th page in some abelian category $\mathcal{A}$. Each page is a subquotient of the next page, so using the standard correspondence theorem for subobjects (which holds in any abelian category), we obtain subobjects $$ 0=B^a\subseteq\dotsc\subseteq B^r\subseteq B^{r+1}\subseteq\dotsc\subseteq Z^{r+1}\subseteq Z^r\subseteq\dotsc\subseteq Z^a=E^a, $$ such that $Z^r/B^r\cong E^r$ in a compatible manner for all $r\ge a$ (I suppress the bigrading notationally). Now, we want to define $B^{\infty}=\bigcup_{r\ge a}B^r$, $Z^{\infty}=\bigcap_{r\ge a}Z^r$ and the limit page to be $E^{\infty}=Z^{\infty}/B^{\infty}$. Weibel points out that unions/intersections of subobjects need not exist in general abelian categories and this is true, but can easily be fixed by assuming that $\mathcal{A}$ satisfies (AB3) and (AB3*) (we could even get away with assuming just the existence countable (co-)products, too). However, Weibel suggests that we should assume (AB4) and (AB4*) instead, which at that point is not intuitive to me.
Proceeding onwards, the first hurdle seems to be the Comparison Theorem 5.2.12. Since conventions are not universal, let me recall that a spectral sequence $E$ converges to a graded object $H$ if there is an exhaustive, complete filtration $F_{\bullet}H$ on $H$ s.t. $E_{pq}^{\infty}\cong F_pH_{p+q}/F_{p-1}H_{p+q}$ for all $p,q\in\mathbb{Z}$ (for example, in Boardman's classical paper, this is called strong convergence instead). Now, suppose there is a morphism $f\colon E\rightarrow E^{\prime}$ of spectral sequences converging to $H$ and $H^{\prime}$ respectively as well as a morphism $h\colon H\rightarrow H^{\prime}$ compatible therewith (meaning the morphisms $E_{pq}^{\infty}\rightarrow E_{pq}^{\prime\infty}$ induced by $f$ and the morphisms $F_pH_n/F_{p-1}H_n\rightarrow F_pH_n^{\prime}/F_{p-1}H_n^{\prime}$ induced by $h$ agree under the isomorphism implied by convergence). The theorem states that if $f^r\colon E^r\rightarrow E^{\prime r}$ is an isomorphism for some $r$, then $h$ is an isomorphism.
I can follow the proof up to the point where it is deduced that $h$ induces an isomorphism $F_pH_n/F_sH_n\rightarrow F_pH_n^{\prime}/F_sH_n^{\prime}$ for all $p\ge s$. Then, Weibel claims that $h$ induces an isomorphism $\bigcup_{p\ge s}F_pH_n/F_sH_n\rightarrow\bigcup_{p\ge s}F_pH_n^{\prime}/F_sH_n^{\prime}$, which (by exhaustiveness) is an isomorphism $H_n/F_sH_n\rightarrow H_n^{\prime}/F_sH_n^{\prime}$. Taking the limit would finish the proof. The scenario is: $$\require{AMScd} \begin{CD} 0=F_sH_n/F_sH_n @>{\subseteq}>> F_{s+1}H_n/F_sH_n @>{\subseteq}>> \dotsc @>{\subseteq}>> \bigcup_{p\ge s}F_pH_n/F_sH_n @>{\subseteq}>> H_n/F_sH_n\\ @VV{\wr}V @VV{\wr}V @. @VVV @VVV \\ 0=F_sH_n^{\prime}/F_sH_n^{\prime} @>{\subseteq}>> F_{s+1}H_n^{\prime}/F_sH_n^{\prime} @>{\subseteq}>> \dotsc @>{\subseteq}>> \bigcup_{p\ge s}F_pH_n^{\prime}/F_sH_n^{\prime} @>{\subseteq}>> H_n^{\prime}/F_sH_n^{\prime} \end{CD}.$$ (I should point out that taking unions/intersections does, in general, commute with taking quotients by the correspondence theorem, so the notation is not ambiguous.) The crux now has nothing to do with spectral sequences anymore and is simply whether, in such a diagram, the isomorphisms to the left imply that the morphism induced between the unions is an isomorphism. I'm only able to argue it is an epimorphism. It would be easily true if the union of an increasing sequence of subobjects could be realized by its colimit, but the natural condition ensuring that would be (AB5) and not (AB4).
Note that a similar question was asked before, yet received no answer (the OP also ends up defaulting to (AB5)). Similarly, user @FShrike recently asked a question on MO dealing with a claim later in the chapter and also arrives at (AB5) being the natural hypothesis to make it work, as opposed to the (AB4) claimed by Weibel (however, no necessity was established).
Thus, my question(s) are:
Is Weibel correct in only requiring (AB4) to make everything work or do we need (AB5) (and what about the (AB4*) condition, which is even more confounding)? If this is incorrect, it would be worth recording with a counter-example.
Is it possible to fix the potential gap in the proof of Theorem 5.2.12 assuming only (AB4)? (If not, this immediately settles question 1. as well. If yes, this might help understanding how to arrive at a positive answer to 1..)
In short: The assumptions Weibel makes do not suffice. Here is an incomplete guide on what goes wrong in generalizing chapter $5$ to arbitrary abelian categories and what can be fixed.
The axioms (AB3), (AB3*), (AB4) and (AB4*) in the following can be weakened to their countable counterparts. The axiom (AB5) can be weakened to the exactness of filtered colimits of $\mathbb{N}$-shaped diagrams.
The axioms (AB3) resp. (AB3*) are necessary to talk about unions resp. intersections of subobjects in general. In the following, it will always be the case that they are either assumed for that reason or that a certain boundedness condition forces the boundary resp. cycle subobjects to stabilize so that the union resp. intersection exists for trivial reasons.
The comparison theorem 5.2.12. needs the assumptions (AB5) and (AB3*) in general. The conditions can be dropped respectively if the spectral sequences are bounded above resp. regular.
Indeed, placing the filtration quotients of a completely, exhaustively filtered object on the diagonal of every page and taking trivial differentials yields a spectral sequence converging to that very object, so we see the comparison theorem implies (and is implied by, as the argument in the post shows) the claim that a morphism of two completely, exhaustively filtered objects inducing isomorphisms on all filtration quotients is an isomorphism.
Here is a counter-example to this in a category satisfying (AB4) and (AB4*), but not (AB5), namely consider the diagram in $\mathbf{Ab}^{op}$ $$\require{AMScd} \begin{CD} \bigoplus_{i\ge0}\mathbb{Z} @>>> \dotsc @>>> \bigoplus_{i=0}^{n+1}\mathbb{Z} @>>> \bigoplus_{i=0}^n\mathbb{Z} @>>> \dotsc @>>> \mathbb{Z} @>>> 0\\ @VVV @. @VV{\wr}V @VV{\wr}V @. @VV{\wr}V @. \\ \prod_{i\ge0}\mathbb{Z} @>>> \dotsc @>>> \prod_{i=0}^{n+1}\mathbb{Z} @>>> \prod_{i=0}^n\mathbb{Z} @>>> \dotsc @>>> \mathbb{Z} @>>> 0 \end{CD}.$$ The horizontal maps are canonical projections and epimorphic. The vertical maps to the right are isomorphisms (and hence, in particular, induce isomorphisms of the filtration quotients). The rows correspond to bounded below (hence complete) filtrations of the leftmost objects, which are exhaustive as the maps $\bigoplus_{i\ge0}\mathbb{Z}\hookrightarrow\prod_{i\ge0}\mathbb{Z}\hookrightarrow\prod_{n\ge0}\prod_{i=0}^n\mathbb{Z}$ are injective. The leftmost vertical map, however, is clearly not an isomorphism. (Note also that the union in the bottom row agrees with the filtered colimit, the union in the top row does not.)
The construction of the spectral sequence of a filtered complex (section 5.4) works in any abelian category, no significant modification required.
The classical convergence theorem 5.5.1 holds without any additional assumptions in the case of a (degree-wise) bounded filtration. In the case of a (degree-wise) bounded below and exhaustive filtration, the axiom (AB5) is necessary. The later applications of this theorem all need to be updated accordingly.
The place we need (AB5) is in the formula $(\bigcup_iA_i)\cap B=\bigcup_i(A_i\cap B)$ for an increasing chain $A_0\subseteq\dots\subseteq A_i\subseteq\dotsc\subseteq A$ of subobjects and a subobject $B\subseteq A$. This is needed to argue that the induced filtration on the homology is exhaustive. In fact, if you consider the complex $C=[A\rightarrow A/B]$ (concentrated in degrees $1$ and $0$), it is exhaustively filtered by the subcomplexes $F_iC=[A_i\cap B\rightarrow A_i/A_i\cap B]$, so the classical convergence theorem implies that $H_1(C)=B$ is exhaustively filtered by $F_iH_1(C)=A_i\cap B$, so these are equivalent.
Here is a counter-example to this in, again, $\mathbf{Ab}^{op}$. As before, consider the exhaustively filtered object $$\require{AMScd} \begin{CD} \prod_{i\ge0}\mathbb{Z} @>>> \dotsc @>>> \prod_{i=0}^{n+1}\mathbb{Z} @>>> \prod_{i=0}^n\mathbb{Z} @>>> \dotsc @>>> \mathbb{Z} @>>> 0 \end{CD}.$$ The subobject $\prod_{i\ge0}\mathbb{Z}\twoheadrightarrow\mathrm{colim}_m\prod_{i\ge m}\mathbb{Z}\neq0$ (the group of "tail ends" of integer sequences) is non-trivial, but the induced filtration on it is trivial as we have pushout squares $$\require{AMScd} \begin{CD} \prod_{i\ge0}\mathbb{Z} @>>> \prod_{i=0}^n\mathbb{Z}\\ @VVV @VVV\\ \mathrm{colim}_m\prod_{i\ge m}\mathbb{Z} @>>> 0 \end{CD}.$$
Lemma 5.5.7, which is the starting point of investigating weak convergence, similarly should require (AB5) and (AB3*), which can be dropped respectively if the filtration is bounded above resp. below.
The condition (AB4*) enters the picture from corollary 5.5.8 onward as Weibel uses the theory of "derived" inverse limits in (AB4*) categories. The proof of the complete convergence theorem 5.5.10 appears to rely on proposition 3.5.7, which states that $\lim^1$ vanishes for a tower satisfying the Mittag-Leffler condition. This was a "theorem" for over 40 years (including at the time Weibel's book was published), however it is incorrect in general, see this MO post and the linked papers.
I see no indication that Weibel's proof could be made to work in general (AB4*) categories. It's also worth noting that Boardman's original paper only worked in the category of abelian groups. The false "theorem" has corrected versions under more rigid hypotheses that probably do imply a complete convergence theorem under these hypothesis. I am out of my depth regarding this theorem, so corrections are welcome.