Given a category $\mathbf{C}$, we can consider monomorphisms into it. These are the faithful and injective-on-objects functors (this violates the [principle of equivalence](https://ncatlab.org/nlab/show/subcategory#VariantsInAccordWithThePrincipleOfEquivalence)). The idea is to try to get a lattice of subobjects.


Pullbacks (in the 1-categorical sense, which [also has problems](https://math.stackexchange.com/questions/168276/pullbacks-of-categories) with principle of equivalence) preserve monomorphisms, so we can have a notion of "meet". Then, given monomorphisms $\mathbf{X} \to \mathbf{C}$ and $\mathbf{Y} \to \mathbf{C}$, we can take the image of the coproduct $\mathbf{X} + \mathbf{Y} \to \mathbf{C}$ and have a notion of "join".
Here the image is the smallest subcategory containing all the images of morphisms under the functor (again, not really treating $\mathbf{Cat}$ as a 2-category).


Is this construction plausible or is there any problem I am overlooking? Does that poset have any extra structure? I am not sure if I should understand from [here](https://ncatlab.org/nlab/show/poset+of+subobjects) that for it to be a distributive lattice we would need $\mathbf{Cat}$ to be coherent, but it is not even regular. If not, can we impose some conditions on $\mathbf{C}$ to get some structure?



Structure of a poset of subcategories
Like Kevin said in the comments, your definition of join should work just fine in any category with finite coproducts and a good notion of "image." The issue with $\textbf{Cat}$ is that you won't have a *distributive* lattice (which is specifically what the nLab says a coherent category will give you).


Here's an example of distributivity failing: Let $\mathcal{C}$ be a category with three objects $x,y,z$ and three non-identity morphisms, $f\colon x\to y$, $g\colon y\to z$, and $h=g\circ f$. By abuse of notation, I'll refer to subcategories by the objects or morphisms which generate them. If we take the join of $f$ and $g$ we get all of $\mathcal{C}$, and so taking the meet of $h$ with that we get $h$. But, if we instead take the meet of $h$ with $f$ and $g$ individually first, we get $x$ and $z$, and then the join of $x$ and $z$ is $x\cup z$, which does not include $h$.


