Let $X$ be a Kolmogorov space in which every non- empty open set is infinite. Show that there exists a quasi-maximal topology on $X$ which is finer than the given topology. (show that the set of topologies on $X$ in which all the non-empty open sets are infinite is inductive.)
I used that if given set is infinite, then $X$ has no isolated point. But how does maximality forces a finer topology than given topology?
Let $\mathcal{P}$ be the poset of all topologies on $X$, ordered by inclusion, that are at least as big (fine) as $\tau_X$ (the given topology) and that also have the property that all open sets are infinite (so that in particular those topologies have no isolated points).
Let $\mathcal{C}$ be a chain in $\mathcal{P}$.
Then let $\tau$ be the topology generated by $\bigcup \mathcal{C}$ (so take that collection of sets as its subbase). Then this topology trivially contains $\tau_X$ too (as all members of $\mathcal{C}$ do) and $\tau$ has no finite open set: suppose that it had, then there would be finitely $O_1,\ldots, O_n \in \bigcup \mathcal{C}$ such that $O:=O_1 \cap O_2 \cap \ldots O_n$ is finite (as finite intersections from the subbase form a base for $\tau$), with $O_i \in \tau_i \in \mathcal{C}$ and as $\mathcal{C}$ is a chain in our poset, there is one $\tau_j \in \{\tau_1,\ldots,\tau_n\}$ that contains all of the $O_i, i=1,\ldots,O_n$. But then $\tau_j$ would have already contained the finite open set $O$ which is not the case (as it is in $\mathcal{P}$). So $\tau$ is a member of $\mathcal{P}$ and an upperbound for the chain $\mathcal{C}$. So $\mathcal{P}$ is inductive.
Zorn's lemma now immediately implies that $\mathcal{P}$ has a maximal element $\tau_{\text{max}}$ and this is the required quasi-maximal topology finer than $\tau_X$.