Let $X\neq \emptyset$ be a set and let ${\cal J} \subseteq {\cal P}(X)\setminus\{\emptyset\}$ be a collection of non-empty subsets of $X$. We say that a topology $\tau$ on $X$ is ${\cal J}$-*compatible* if for every $J\in {\cal J}$ there is a continuous surjective map $f:X\to J$, where $J$ carries the subset topology inherited from $(X,\tau)$.


Let $\text{Cptb}(X, {\cal J})$ denote the collection of ${\cal J}$-compatible topologies, ordered by $\subseteq$.


Note that ${\cal P}(X)$ is always the greatest element of $\text{Cptb}(X,{\cal J})$, and the indiscrete topology $\{\emptyset, X\}$ is always the smallest element. 


If $\tau\in \text{Cptb}(X, {\cal J})$ and $\tau\neq{\cal P}(X)$, is there a minimal element in $\text{Cptb}(X, {\cal J})$ properly containing $\tau$ (= "above" $\tau$, in the poset we are considering)?



Partially ordered set of compatible topologies
If we put $\mathcal J = \{X\}$, then $\mathrm{Cptb}(X,\mathcal J)$ is just the lattice of topologies on $X$. In this case, the answer to your question is known: if $X$ is infinite, then not every topology on $X$ has a "minimal" strict refinement.


For example, let $p \in X$ and let $\tau$ be the one-point compactification of $X - \{p\}$: that is, neighborhoods of $p$ are sets of the form $X - F$, where $F$ is a finite subset of $X$ not containing $p$, and every other point is isolated. I claim that $\tau$ has no minimal proper refinement. To see this, suppose $\tau'$ properly refines $\tau$. Then there is some infinite closed $Y \subseteq X - \{p\}$. Split $Y$ into two disjoint infinite sets $Y\_0$ and $Y\_1$, and define $\sigma$ to be the refinement of $\tau$ obtained by declaring $Y\_0$ closed (in other words, $\sigma$ is the topology with subbasis $\tau \cup \{X - Y\_0\}$). It is not hard to see that $\sigma$ is strictly between $\tau$ and $\tau'$.


