An ordered set can be regarded as a category. What are its subcategories? Which of them are full?
Here's what I understand.
The elements of the ordered set correspond to the objects of the category. There are at most one arrow between any two objects, and there is an arrow $a\to b$ iff $a\le b$ in the ordered set. A subcategory consists of a subset of object and a subset of arrows subject to some conditions. So at least a subcategory is a subset of the ordered set. It also comes with a collection of arrows. For any two objects $a,b$ in the subcategory, if there is no arrow from $a$ to $b$ in the original category, then there cannot be arrow in the subcategory. If there is an arrow $a\to b$ in the original category, then this arrow may or may not belong to the subcategory. In particular, it cannot be the case that there is an arrow $a\to b$ in the subcategory but there is no arrow $a\to b$ in the original category. If $a\to b$ and $b\to c$ are in the subcategory, then $a\to c$ must be in the subcategory. But I don't see how to extract a reasonable classification of subcategories (including full subcategories) from all of the above.
There is a question that includes my question: What are the subcategories of ordered sets / groups?
But I don't understand the answer (see my comment to the accepted answer and see the above for an explanation/proof of the claim of the comment).
This is a little bit expanded version of this answer; thanks to Giorgio Mossa for the discussion in the comments.
Let $\mathscr P$ be a category corresponding to a poset $(P,\le_P)$. Consider a subset $Q\subseteq P$ of the set of objects of $\mathscr P$. Assume it is the set of objects of a subcategory $\mathscr Q$ of $\mathscr P$. Since there are at most one arrow between any two objects of $\mathscr P$, there is also at most one arrow (in $\mathscr Q$) between any two elements of the subset $Q$ we are considering. Define the relation $\le_Q$ on the set $Q$ as follows: $a\le_Q b$ iff there is an arrow from $a$ to $b$ in $\mathscr Q$. We claim that $\le_Q$ is a partial order.
Since $\mathscr Q$ contains all identity arrows for every object of $\mathscr Q$, the relation $\le_Q$ is reflexive. It is also transitive since $\mathscr Q$ is closed under composition. The anti-symmetry of $\le_Q$ follows because if $a\to b$ and $b\to a$ in $\mathscr Q$ with $a\ne b$, then we also have $a\to b$ and $b\to a$ in $\mathscr P$ with $a\ne b$, which contradicts the fact that $\le_P$ is anti-symmetric. These three facts show that $\le_Q$ is a partial order.
Now note that if there were no arrow $a\to b$ in $\mathscr P$, then there would not be such an arrow in $\mathscr Q$ either. Taking the contrapositive and translating into the language of orders, this means that if $a\le_Q b$, then $a\le_P b$.
Therefore, the subcategories of $\mathscr P$ are the categories corresponding to a poset $(Q,\le_Q)$ with the property that for all $a,b\in Q$, one has $a\le_Q b\implies a\le_P b$, together with the empty subcategory.
A subcategory $\mathscr Q$ of $\mathscr P$ is full if for all $a,b\in Q$, if $a\to b$ in $\mathscr P$, then also $a\to b$ in $\mathscr Q$. In the language of orders, this means that for all $a,b\in Q$, if $a\le_P b$, then $a\le_Q b$. Thus the full subcategories are those corresponding to a poset $(Q,\le_Q)$ with the property that for all $a,b\in Q$, one has $a\le_Q b\iff a\le_P b$, together with the empty subcategory.