Order Properties on Open Sets Considering the subset order on the open sets of a topological space, it seems natural to ask what kind of total orders exist as suborders of the subset order. One possibility is that each total order is a well order. Where can I read more about this kind of thing?  
 A: In point set topology, one can define the open sets $\mathcal{T}$ of a topological space $(X,\mathcal{T})$ however one wants as long as the elements of $\mathcal{T}$ have the following three properties:  First, both $\varnothing$ and $X$ must be open. Second, the intersection of two open sets must also be open.  Third, the union of any collection of open sets is open.
One can translate these conditions to the language of ordered sets.  Clearly, the subset order on the open sets of a topological has a maximal, has a minimum, has finite infimums, and has arbitrary supremums.  In other words, such an ordered set is a bounded lattice that happens to be closed under arbititrary supremums / 'meets'.  
Conversely any lattice $L$ of this sort is actually the set of open sets of a topological space $(L, \mathcal{L})$ as follows:  consider all subsets $\mathcal{L}$ of $L$ that consist of some element $l \in L$ along with all elements $k \in L$ such $k \leq l$.  This class of sets is closed under arbitrary intersection, arbitrary union, and contains both $\varnothing$ and $L$ and thus forms the desired topology.
Going back to your question about total suborders of the open set order: any any total order $T = (|T|, \leq_T)$ is  a suborder of another total order with both a 'bottom' element (like $\varnothing$) and a 'top' element (like $K$).  This order is, in turn, a supremum closed lattice which is isomorphic to some 'open sets in a topology' order by the paragraph above.
A related question might be which (total?) orders show up in the open subset order of a metric space.  Or, for that matter, any restricted class of topological spaces.  I believe one can say this:  A total order is not the suborder of an open-set order unless it countable.
A: This answer is partially based on two comments to the question, one by @BrianM.Scott and one mine. It consists of two observations.

Proposition 1. Every total order $(X,\leq)$ embeds into a $(\mathcal{T},\subseteq)$ for some topology $\mathcal{T}$.

This can be seen by taking $\mathcal{T}$ to be the topology on $X$ with base
$$
\{I(a,b) : a< b \},
$$
where $a,b\in X$ and $I(a,b) := \{x \in X : a<x<b\}$ is the open interval with endpoints $a$ and $b$. This construction, called the order topology on $X$, is totally analogous to that of the topology on $\mathbb{R}$.
Now it is easy to see that the map 
$$
x\mapsto \{y \in X : y< x\}
$$
is an order embedding.

Proposition 2. Let $(X,\mathcal{T})$ be a $T_1$ topological space. If every total suborder of $(\mathcal{T},\subseteq)$ is a wellorder, then $X$ is finite.

Here, $T_1$ means that singletons are closed (this happens in every Hausdorff space, for example). To prove the Proposition, we show the contrapositive. If $X$ is infinite, take some infinite sequence $\{x_n: n\in \mathbb{N}\}\subseteq X$ and consider the open sets
$$
X \supset X\setminus\{x_1\} \supset X\setminus\{x_1,x_2\}\supset \dots
$$
This is a strictly decreasing $\subseteq$-sequence, hence not a wellorder.
