If an abelian category has a generator then it is well-powered The title of my question is Proposition 3.35 in Freyd's Abelian categories. The proof says

If $G$ is a generator and $A$ is any object, then a subobject $A' \longrightarrow A$ is distinguished by the subset $(G,A') \subseteq (G, A)$.

What does "distinguished by" mean and why does this prove the proposition?
 A: By "generator" Freyd surely means "strong generator". Anyway, let $A' \to A$ be a monomorphism, and consider the map $\textrm{Hom}(G, A') \to \textrm{Hom}(G, A)$. This is certainly injective, by the definition of monomorphism, so we may identify $\textrm{Hom}(G, A')$ with a subset of $\textrm{Hom}(G, A)$. 
Suppose $A'' \to A'$ is another monomorphism. Then we get a chain of subsets
$$\textrm{Hom}(G, A'') \subseteq \textrm{Hom}(G, A') \subseteq \textrm{Hom}(G, A)$$
and the definition of strong generator implies that $\textrm{Hom}(G, A'') = \textrm{Hom}(G, A')$ if and only if $A'' \to A$ and $A' \to A$ are isomorphic as subobjects of $A$. But what if $A' \to A$ and $A'' \to A$ are not comparable? Well, then take their pullback (which exists in any abelian category) and then use that as a bridge to compare $A' \to A$ and $A'' \to A$. The conclusion is, $\textrm{Hom}(G, A') = \textrm{Hom}(G, A'')$ as subsets of $\textrm{Hom}(G, A)$  if and only if $A' \to A$ and $A'' \to A$ are isomorphic as subobjects of $A$.
In particular, since $\textrm{Hom}(G, A)$ only has a set of subsets, $A$ only has a set of (isomorphism classes of) subobjects.
A: I'm not sure if the answer below is more well-suited to be a comment or as an answer, but I think that it might be helpful to include a short proof that in the setting of Freyd's book that the assertion can be proved directly without assuming that $G$ is strong. (but may only work in abelian categories)

Case 1 : Comparable Objects
By the definition of a generator (not necessarily strong), we see that the functor $\operatorname{Hom}(G,-)$ is left exact and faithful. For clarity, let us call this functor $F$.
Suppose we are given a proper inclusion $A'\rightarrow A$, we may extend the inclusion to an exact sequence $0\rightarrow A'\rightarrow A\rightarrow A''\rightarrow 0$ and apply the functor $F$.
If we were to have an exact sequence $0\rightarrow FA'\rightarrow FA\rightarrow 0$ then one concludes by faithfulness of $F$ that $A\rightarrow A''$ is just $0$ and get a contradiction. This shows that $FA'$ is a proper subset of $FA$ whenever $A'$ is a proper subset of $A$.
Case 2 : Incomparable Objects
This is done in Zhen Lin's answer.
(To be more precise : one can show that $F$ "commutes with" $\cap$ in some sense)
