Your claim that those hypotheses ensure that $I$ is a topology is not correct. What you have is a family of subsets of $X$ that is linearly ordered by $\subset$ and includes the empty set and $X$ itself, and not every such family is a topology. For example, consider the family of intervals in the real line of the form $(-q,q)$, for $q\in\mathbb{Q}$, plus the empty set and all of $\mathbb{R}$. These intervals are nested in the sense you describe, but they do not form a topology, since this family is not closed under arbitrary unions.

Meanwhile, if you have an actual topology that consists of a family of sets that is linearly ordered by $\subset$, then this topology is compact if and only if it contains a largest proper subset of $X$. If it does have such a set, then every open cover must contain the whole set $X$, since the union of all smaller sets does not cover the space. Conversely, if it does not have such a set, then the union of all the proper subsets of $X$ is $X$ itself, and so this will be an open cover with no finite subcover.

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