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As stated, this looks like a yes/no question, as in a poll. As such, it doesn't seem like an ideal question for MathOverflow. Since it looks like a poll, I'll make it Community Wiki (but the community may decide it's not quite right for this site).
It looks as though the MSE bounty is expired. I could help (over the next few days), but I think I'd prefer to post over there. The problem may be that full details are somewhat tedious; this may account for lack of response.
If you're not referring to the extension of scalars functor, tensoring with $T$ over $S$, then I don't know what else you'd be referring to. On the other hand, for that to be fully faithful doesn't seem too common. So I'm not sure. I think maybe you should think about it some more, since you know better what the outlying context is.
The question formulation seems odd to me because $F$ is just any old functor, not connected to the hypotheses like $S \subset T$ being faithfully flat. But a fully faithful functor $F: C \to D$ is an equivalence iff it is essentially surjective on objects, where "essentially surjective" means every object $d$ of $D$ is isomorphic to $F(c)$ for some object $c$ of $C$. If $F$ is essentially surjective and if I have understood your notation, then $F \otimes 1$ is essentially surjective as well, so (1) implies (2).
We're glad you're interested in mathematics, and hope you keep it up. For the most part, MO is for professionals and the community expects answers at a professional level, and this answer doesn't quite reach that level, hence the down-voting and votes to delete. Don't take it too much to heart, but for future reference, please be aware of the level.
@Theo111 Oh, it's very clear why that's very confusing -- sorry! What I was really trying to get at was the fact that if you have a subobject in this category, i.e., an equivalence class of monomorphisms $A \hookrightarrow X$, then the topology on $A$ must be given by the subspace topology. This is not true in $Top$ for example, as I indicated in my answer. Of course it's ridiculous to say that all subspaces of a compact Hausdorff space are compact Hausdorff, and that was one way of interpreting my less than optimal phrasing. Anyway, closed subspaces are the same as subobjects, whew!
