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, 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.