Open subset is compact iff it is finite Let $X$ be a discrete space. Then, obviously, an open subset of $X$ is compact if and only if it is finite. Are there any other topological settings (coarse enough to be non-trivial) for infinite topological spaces to have this property?
 A: As noted in my comment above, in quite a few topologies, there are no finite open sets and no compact open sets. In these cases, your statement is vacuously true.
For example, $X=\mathbb R$ with the standard topology has your property, because no open sets are finite, and no open sets are compact.
The rest of my examples will have cases with finite open sets.
Let $X=\mathbb N$ with the topology where the open sets $U$ have the property that $n\in U$ and $m\leq n$ implies $m\in U$. This has your property, since the only infinite open set is all of $X$, and it is not compact.
Given a topology $\tau$ on $X$, and an $x\in X$, define a topology $\tau_x$ which includes:
(1) Any subset of $X\setminus\{x\}$ and
(2) Any superset of a $U\in \tau$ such that $x\in U$.
Then, in certain circumstances, I think $\tau_x$ will have the property you want. (I think the property is that for any $U\in \tau$ with $x\in U$, there exists $V\in\tau$ with $x\in V$ such that $U\setminus V$ is infinite.)
This space is "close to discrete," in that it is discrete on $X\setminus\{x\}$ but more complicated around $x$. (It can be thought of as encoding the idea of "continuity at $x$.") It is Hausdorff when $\tau$ is Hausdorff (since $\tau\subseteq \tau_x$).
