$\DeclareMathOperator{\colim}{colim}$Let $I$ be a (small) category, and take $S(I)$ to be the poset of finite subcategories of $I$, ordered by inclusion, treated as a category. One can show $S(I)$ turns out to be filtered (EDIT: not true, see below). This gives a functor $S(I)\to Cat$ sending a subcategory $I'$ to itself seen as category in its own right. We can look at the colimit of that functor (assuming it exists, I think it does, thanks to some result I do not know about small colimits in $Cat$)
I'm fairly confident that the canonical morphism $f:\colim_{I'\in S(I)}I'\to I$ is an equivalence of categories. This is very reminiscent of say "a module is isomorphic to the filtered colimit of its finite type submodule" or "a set is isomorphic to the filtered colimit of its finite subset". But in that case, I'm not sure how to proceed, because I don't really know how to handle colimits in $Cat$, my intuition breaks down so to speak.
I think I can show $f$ to be essentially surjective (see below), but I don't know how to proceed onwards and prove it's fully faithful. Producing a functor in the other direction and the required natural isomorphisms seems quite a bit harder, so I'd rather stick to showing it's fully faithful.
For the essential surjectivity, take $A\in I$ then $A\in I'$ for some finite subcategory of $I$, hence we get an objet of $\colim_{I'\in S(I)}I'$ by looking at the image of $A$ via the canonical morphism $\iota_{I'}:I'\to\colim_{J\in S(I)}J$, i.e. $\iota_{I'}(A)$. We then have $A\simeq f(\iota_{I'}(A))$.
It would now suffice to show $\hom_{\colim_{I'\in S(I)}I'}(A,B)\simeq \hom_{I}(fA,fB)$. But I have no idea what the first hom-set looks like.
EDIT : As @AlexKruckman pointed out in the comments, as stated, there is an issue. I said at the beginning "One can show $S(I)$ turns out to be filtered", but this is not the case. As he mentioned, this fails to be true for instance when considering a free product $G*H$ with $G,H$ finite and non trivial. So one should replace "finite" with finitely generated here.