OK, I am looking for no "hands waiving" proof that inverse limit over a cofinal index subset is isomorphic to the inverse limit over its superset. For instance: Given an (Abelian) category $C$ with infinite products (a sufficiently normal one), and an inverse system $f_{ij}:A_j\longrightarrow A_i$ indexed by an upward directed set $I$ that has a cofinal subset $J$, prove that $\varprojlim_I A_i\cong\varprojlim_J A_j$. I do not assume $J$ to be linearly ordered, rather it only satisfies $J\subseteq I$ and $\forall i\in I\exists j\in J$ with $i\leq j$. If $J$ is assumed to be linearly (hence) well ordered, the proof is not too hard.
[Math] Inverse limits over cofinal subsets
category-theorylimits-colimits
Related Solutions
You cannot do this if you want each of the objects and maps in your new system to be objects and maps of the old system. For instance, if start with any inverse system of finite sets in which all the maps are surjections, this would imply that the ordinal-indexed system is essentially countable (i.e., if you discard maps which are isomorphisms, there are only countably many terms, so that there is a countable subsequence with the same limit), since the cardinality of $X_\alpha$ can increase only countably many times. That would in particular imply the cardinality of the inverse limit is at most $2^{\aleph_0}$, since it is a countable inverse limit of finite sets. But every infinite set can be written as an inverse limit of finite sets and surjections (see https://mathoverflow.net/questions/172707/maximum-cardinality-of-a-filtered-limit-of-finite-sets).
If you don't mind modifying the objects, then there are some constructions you can make. For instance, let us suppose $J$ has finite joins (if it doesn't, you can formally adjoin them, and extend the system by mapping the formal join of $F\subseteq J$ to the limit of the diagram of elements that are below some element of $F$; alternatively, instead of using joins we can just arbitrarily pick some upper bounds in the construction below, with a bit of extra work). Fix an enumeration $(j_\alpha)_{\alpha<\kappa}$ of $J$ indexed by some ordinal $\kappa$. For each $\alpha<\kappa$, let $X_\alpha$ be the limit of the diagram formed by the objects $X_s$ where $s$ ranges over the set of finite joins of elements $j_\beta$ for $\beta<\alpha$ (when $\alpha$ is finite, this just means $X_{\alpha}=X_s$ where $s$ is the join of $j_0,\dots,j_{\alpha-1}$). Then the $X_\alpha$ naturally form an inverse system whose inverse limit is the same as the inverse limit of the original system.
Note in particular that if we pick $\kappa$ to be a cardinal, then each $X_\alpha$ is the inverse limit of an inverse system of smaller cardinality. So by induction on $\kappa$, this iteratively constructs our inverse limit out of ordinal-indexed inverse limits (a countable inverse limit is an ordinal-indexed limit built from the original diagram, a size $\aleph_1$ inverse limit is an ordinal-indexed limit where each term is a countable inverse limit built from the original diagram, etc).
This construction and variants on it are useful for some purposes; for instance, it can be used to prove that if a functor preserves ordinal-indexed limits (and those limits always exist) then it preserves all inverse limits.
I think the idea is good; let me comment on how I would think about it:
In order for the inverse limit $\varprojlim_{i \in I} X_i$ to be defined, you need $I$ to have the structure of a category, and you need $i \mapsto X_i$ to have the structure of a functor. You do this by making $I$ into a preorder when you define the "$\leq"$ relation, but to be super-pedantic, you didn't explicitly check that the relation $\leq$ is reflexive and transitive. Then you do check that $i \mapsto X_i$ is a functor with the action $(i\leq j) \mapsto \iota_{ij}$ on morphisms. To be super-pedantic again, you never explicitly said that $X$ was to be a functor to the category $Set$ (rather than to some other category); I suppose this was understood, but see the second bullet below.
You then define a cone on the functor $X: I \to Set$, with vertex $\cap_{i \in I} X_i$ and legs $\iota_i$, and verify that this is indeed a cone. Great.
Next you verify the universal property, which I think looks good modulo the one significant issue of your post which I come to next.
The only real issue is the following. In the hypotheses, you never stipulated that the family of sets $(X_i)_{i \in I}$ be downward directed in the sense that for every $i,j \in I$ there exists $k$ with $X_k \subseteq X_i \cap X_j$. But you use this assumption in your proof of the universal property, when you produce the index $k$.
So what you've shown is that if $(X_i)_{i \in I}$ is a family of subsets of $X$ which is downward-directed, then $\varprojlim_{i \in I} X_i = \cap_{i \in I} X_i$. This is good. Without the hypothesis that $(X_i)_{i \in I}$ be downward-directed, the statement would be false in general. For example, suppose that $I = \{1,2\}$ and neither $X_1$ nor $X_2$ is contained in the other. Then $I$ is a discrete poset with the $\leq$ order you define, and $\varprojlim_{i \in I} = X_1 \times X_2$, which of course is typically different from $X_1 \cap X_2$.
Some related notes:
In the usage I'm familiar with, the terms "inverse limit" and "projective limit" are ambiguous. Sometimes they mean (1) "limit indexed by a downward-directed poset" (or more generally, "limit indexed by a co-filtered category"), but sometimes they mean more generally (2) "limit indexed by an arbitrary category". I think that at least nowadays most people tend to just say "limit" when they mean (2), so that the usage (1) is more common, but it's something to watch out for.
In general the limit $\varprojlim_{i \in I} X_i$ can change depending what category you regard the assignment $i \mapsto X_i$ as taking values in. Let's go back to the example where $I = \{1,2\}$ is discrete. If you regard the assignment $i \mapsto X_i$ as taking values in the category $Set$, then as remarked above, you get $\varprojlim_{i \in I} X_i = X_1 \times X_2$, the cartesian product. But instead, you might regard $i \mapsto X_i$ as a functor $I \to Set/ X$, where $Set / X$ is the slice category, i.e. the category where an object is a set $Y$ equipped with a map $Y \to X$, and a morphism is a map $Y \to Y'$ making the obvious diagram commute. If you do this, then you do end up getting $\varprojlim_{i \in I} X_i = X_1 \cap X_2$. Similarly, if you regard $i \mapsto X_i$ as taking values in the poset $Sub(X)$ of subsets of $X$ and inclusions (note that this is a full subcategory of $Set/X$), then you again get $\varprojlim_{i \in I} X_i = X_1 \cap X_2$.
Similarly, the limit can change depending on what we regard the morphisms of $I$ as being, so it's good that you were explicit about this. An exception is when we compute limits in a category which is a preorder. In this case, $\varprojlim_{i \in I} X_i$ is the greatest lower bound of the $X_i$'s (either existing if the other does), and it doesn't matter what category structure you regard $I$ as having for this to be true.
Best Answer
For this particular situation, argue as follows. Let $(A,g_j)$ be the limit over $J$; let us show that $(A,f_{ij}g_j)$ is a limit over $I$. First, it's a cone: $f_{jk}f_{ij}g_j=f_{ki}g_j$ by functoriality. Now let $(B,h_i)$ be a cone over $I$. This restricts to a cone $(B,h_j)$ over $J$, and the latter gives rise to a unique $k:B\to A$ with $g_j\circ k=h_j$ for $j\in J$. Now given any $i\in I$, let $j\in J$ and $f_{ij}:A_j\to A_i$. Then $f_{ij}g_jk=f_{ij}h_j=h_i$, which shows the $h_i$ factor through $A$. And this was the only possible factorization $k$, since $k$ is determined even by the restriction to $J$.