A topological space $X$ is compact if and only if any collection of closed sets satisfying the finite intersection property has non-empty intersection.
Clearly, this implies that compact spaces $X$ satisfy the property that any descending chain $\{Y_i\}_{i\in I}$ of closed non-empty sets has non-empty intersection. The converse, that a set $X$ satisfying this condition is compact, seems like it should be false. However, I can't find a counterexample. What is the right way to think about this/go about finding one?
It's in fact true. Suppose $X$ is not compact. Let $\mathscr U=(U_\alpha)_{\alpha<\kappa}$ be an open cover of $X$ with no finite subcover, where the cardinal $\kappa$ is the smallest possible. Noncompactness implies $\kappa \ge \omega$.
For $\alpha<\kappa$, set $$F_\alpha=X\setminus\bigcup_{\beta<\alpha} U_\beta,$$ this forms a decreasing chain of closed sets.
$\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$, because $\kappa$ is a limit ordinal. If $F_\alpha = \varnothing$ for some $\alpha<\kappa$, then $(U_\beta)_{\beta<\alpha}$ is an open cover of $X$ with no finite subcover. But the minimality of $\kappa$ implies $|\alpha|\ge\kappa$, which is a contradiction.
Therefore $(F_\alpha)_{\alpha<\kappa}$ is a decreasing chain of nonempty closed sets, which has empty intersection.