Examples of sets which are not obviously sets

Consider compact metrizable topological spaces, modulo homeomorphism there's actually set many such spaces.

One way to show this is to show that for every compact metrizable space $X$ there is an embedding $X\to [0,1]^\Bbb N$, so there are at most $|\mathcal P([0,1]^\Bbb N)|=2^{|[0,1]^\Bbb N|}=2^\mathfrak c$ such spaces.

Based on the comments (in particular, when I say "isomorphism classes of ... form a set", what I really mean is "there is a set such that any ... is isomorphic to one element of this set"; or I could argue that what I say makes sense using Scott's trick : this all should be equivalent anyway), here are some examples :

1- Given a cardinal $\kappa$, a first order language $L$ and a theory $T$ in $L$, the isomorphism classes of models of $T$ form a set. An interesting special case is when you take $L$ to consist entirely of function symbols and $T$ a list of universally quantified equations; then knowing that this is a set actually helps in one of the usual proofs of the Birkhoff variety theorem.

It's important to note that I restricted to a first order language, but actually there's no need for symbols in $L$ to be finitary, as long as $L$ is a set (and so its symbols have bounded (possibly infinite) arity), this works and the proof is the same.

2- If you include "second countable" in the definition of manifold (I don't know how consensual that is, my teachers did it that way), then diffeomorphism classes of manifolds form a set. This can be interesting to know in some considerations where we have functors defined on $\mathrm{Diff}$, and so to deal with functor categories, knowing that this category is essentially small is interesting (e.g. to see that prestacks on $\mathrm{Dif}$ actually form a category, or cobordism categories)

3- Given a cardinal $\kappa$, homeomorphism classes of compact Hausdorff spaces of cardinality $\leq \kappa$ form a set. This follows because the topology on a compact Hausdorff space is completely determined by the $\lim$ function $\beta X\to X$ ($\beta X$ is the set of ultrafilters on $X$). I don't know how useful that is, but I used it (rather a trivial variation of it) in a bachelor project to prove the existence of a universal minimal flow of a topological group.

4- (part of ) 1- and 3- generalize cleanly to monadic categories : whenever a category of objects is equivalent to $\mathbf{Set}^T$, the Eilenberg-Moore category of $T$-algebras for a certain monad $T$, then for any cardinal $\kappa$ there is a set of isomorphism classes of objects of this category of cardinality $\leq \kappa$. In 1-, $T$ is the monad associated to an algebraic theory, in 4- $T$ is the ultrafilter monad.

One can modify slightly 4 (or 1) to get that isomorphism classes of "finitely generated stuff" usually form a set : $R$-modules for any ring $R$ for instance.

5- Homeomorphism classes of separable metric spaces (in fact, isometry classes of separable metric spaces) form a set. In fact, with the ultrafilter trick as above, given a cardinal $\kappa$, homeomorphism classes of Hausdorff spaces that have a dense subset of size $\leq \kappa$ form a set.

6- As Randall pointed out in the comments, given a model category $C$ and objects $X,Y\in C$ it is not clear at all that $\hom_{\mathbf{Ho}(C)}(X,Y)$ should be a set. In fact if we don't take a model category but just a category with weak equivalences and localize, in general these don't form sets. But in a model category, fibrant and cofibrant replacements and various lemmas ensure that $\hom_{\mathbf{Ho}(C)}(X,Y)$ is (isomorphic to) a quotient of a hom-set in $C$ and is therefore essentially a set.

7- An example that is not "isomorphism classes of", that is also a bit weird, but that has some interesting (philosophical) implications : the class of all sets $x$ such that $\neg\mathrm{(Fermat's\, last\, theorem)}$ is a set (known since the 90's only !)