It is well known that $\mathbf{Set}$ is an $\aleph_0$-accessible category, but I'm very inexperienced and I'm not sure how to prove it in detail. In particular, I need to find a set $\Omega$ of finitely presentable objects (sets) such that every other set is a directed colimit of elements of $\Omega$.
Here is my attempt. Given any set $A$, we can define a directed diagram $([A]^{\aleph_0},\subseteq)$, where $[A]^{\aleph_0}$ denotes the set of finite subsets of $A$. If we define $D:([A]^{\aleph_0},\subseteq)\to\mathbf{Set}$ by $D(S)=S$ and $D(S_1\to S_2)=i_{12}$, where $i_{12}$ is the inclusion from $S_1$ to $S_2$, then $(S\stackrel{i_S}{\longrightarrow} A)_{S\in [A]^{\aleph_0}}$ is a colimit of $D$, where $i_S$ is the inclusion from $S$ to $A$.
In addition, we consider any diagram $F:(I,\leq)\to\mathcal{K}$ with colimit $(F_i\stackrel{\eta_i}{\longrightarrow} \text{colim}( F))_{i\in I}$ and we take some object $C$ isomorphic to $F_j$ (with isomorphism $h:C\to F_j$), for some $j\in I$. Then, the diagram $F':(I,\leq)\to\mathcal{K}$ given by $F'(j)=C$, $F'(i)=F(i)$ if $i\not=j$ and $F'(i\to j)=h^{-1}\circ F(i\to j)$, $F'(j\to i)=F(j\to i)\circ f$ has a colimit $(F_i'\stackrel{\mu_i}{\longrightarrow} \text{colim}( F'))_{i\in I}$ given by $\text{colim}( F')=\text{colim}( F)$ and $\mu_i=\eta_i$ if $i\not=j$ and $\mu_j=\eta_j\circ h$. Basically, we are replacing an object in the diagram and $\text{colim}(F)$ does not change. Is this correct?
Finally, we can replace $D(S)$ with $n$, where $n=|S|$, for every $S\in [A]^{\aleph_0}$. Since $n$ is isomorphic in $\mathbf{Set}$ to any set with cardinality $n$, the previous paragraph implies that the resulting diagram $D'$ (after replacing every object with its cardinality), has $\text{colim}(D')=\text{colim}(D)=A$. Then, this proves that any set $A$ is the directed colimit of sets in $\omega$ (the set of natural numbers), which is waht we wanted to prove.
Is the above reasoning correct? If not, where did I go wrong? And if yes, isn't it a little strange that an uncountable set can be constructed from a diagram with only the natural numbers, which is a countable set?
Yes, you can replace the values of the functor $D$ by the finite cardinals. (Remember that in a colimit diagram, you're allowed to use the same object multiple times!) Thus what you did is correct.
To make the fact that you constructed a way to write an uncountable set as a colimit of finite cardinals more digestable, remember that a directed colimit in $\mathbf{Set}$ is just a directed union, so what we are doing is writing $A=\bigcup_{X \in [A]^{\aleph_0}} X$, i.e. we're just saying that every set is the union of its finite subsets. Now because we can use the same object more than once in the diagram we're taking the colimit over, we can replace this diagram with one that has only objects in $\omega$. If $A$ is uncountable, then the diagram will contain uncountably many copies of $1$, for example.