On the generalized Sierpinski space In Sierpiński topology the open sets are linearly ordered by set inclusion, i.e.  If  $S=\{0,1\}$, then the Sierpiński topology on $S$ is the collection $\{ϕ,\{1\},\{0,1\} \}$ such that
$$\phi\subset\{0\}\subset\{0,1\}$$
we can generalize it by defining a topology analogous to Sierpiński topology with nested open sets on any arbitrary non-empty set as follows:
Let $X$ be a non-empty set and $I$ a collection of some nested subsets of $X$ indexed by a linearly ordered set $(\Lambda,\le)$  such that $I$ always contains the void set $ϕ$ and the whole set $X$, i.e. 
                       $$I=\{\emptyset,A_\lambda,X:A_\lambda\subset X ,\lambda\in\Lambda\}$$
such that   $A_\mu⊂A_\nu$  whenever  $\mu\le\nu$.
Then it is easy to show that $I$ qualifies as a topology on $X$.
My questions are -
(1) Is there a name for such a topology in general topology literature?
(2) Is there any research paper studying such type of compact, non-Hausdorff and connected chain topologies?
 A: This was already pointed out in comments, but I'm collecting this into an answer. (In fact, this is closer to a longer comment than to an answer, but the details can be elaborated better here than in a short comment.)
Both examples here deal with your construction in the special case that $X=\Lambda$. And in both cases we work with the sets of the form
$(-\infty,a)=\{x\in\Lambda; x<a\}$, which is similar to lower topology.

As noticed by Niels Diepeveen, see his comment, this is not necessarily a topology.
Suppose that we have $X=\Lambda$ and $A_\lambda=(-\infty,\lambda)=\{x\in\Lambda; x<\lambda\}$.
This means that for each bounded subset $D\subseteq\Lambda$ there exists a $\lambda$ such that  $\bigcup_{\mu\in D} A_\mu = A_\lambda$. It is relatively easy to show that the last condition is equivalent to $\lambda=\sup_{\mu\in D} \mu$. I.e. the linear order would have to be complete.
A counterexample suggested by Niels is $\Lambda=X=\mathbb Q$. For example if $(q_n)$ is an increasing sequence of rational numbers such that $q_n\to\sqrt{2}$, then $\bigcup (-\infty,q_n)=(-\infty,\sqrt2)$, which is not of the form $A_\lambda$.
There are several possibilities how to circumvent this. E.g. you could take all downward closed sets as open. Or you could take $\{\emptyset,X,A_\lambda; \lambda\in\Lambda\}$ as a base for the topology you want to generate.

Even if this is a topology, it need not necessarily be compact.
Let us take $X=\Lambda=\mathbb Z$.
Again $A_\lambda=\{x\in\mathbb Z; x<\lambda\}=(-\infty,\lambda)\cap\mathbb Z$. I.e. $A_\lambda$'s are down-sets of the linearly ordered set $\mathbb Z$.
Now if $\mathcal C$ is open cover of $X$ then, for each $n\in\mathbb Z$, the cover $\mathcal C$ must contain some $A_\lambda$ with $\lambda>n$. This shows that every open cover is infinite.
And if we take $\mathcal C=\{A_n; n\in\mathbb Z\}=\{(-\infty,n)\cap\mathbb Z; n\in\mathbb Z\}$, we get an open cover which does not have a finite subcover.
