The goal. Let $X$ be a set endowed with Hausdorff topologies $\tau_w$ and $\tau_n$, such that $\tau_w\subseteq\tau_n$. Let $\mathscr{C}$ denote a family of subsets $A\subseteq X$, which satisfies the following properties.
(i) $\mathscr{C}$ is closed under arbitrary intersections and finite unions;
(ii) every $\tau_n$-compact set belongs to $\mathscr{C}$; and
(iii) every $A\in\mathscr{C}$ is $\tau_w$-compact.
I would like to define a new topology $\tau_\mathscr{C}$ on $X$ which satisfies the following properties.
(1) $\tau_w\subseteq\tau_\mathscr{C}\subseteq\tau_n$;
(2) every $\tau_\mathscr{C}$ compact set belongs to $\mathscr{C}$; and
(3) every $A\in\mathscr{C}$ is $\tau_\mathscr{C}$-compact.
Discussion.
The obvious thing to try is to take the intersection $\tau_\cap$ of all topologies $\tau$ satisfying $\tau_w\subseteq\tau\subseteq\tau_n$ and for which every $\tau$-compact set belongs to $\mathscr{C}$. However, it is far from obvious that $\tau_\cap$ would satisfy (2) or (3).
Probably this is not possible in general. However, we could assume that $X$ is a Banach space, $\tau_n$ is the norm topology, and $\tau_w$ is the weak topology. We could also, if necessary, impose some additional assumptions on $\mathscr{C}$.
This seems hopeless in general. Of course if $\tau_w$ and $\tau_n$ have the same compact sets then the only possible $\mathscr{C}$ is this common collection of compact sets and therefore any intermediate $\tau_\mathscr{C}$ will do the job (including $\tau_w$ and $\tau_n$ themselves). However, if there is at least one $A \subseteq X$ which is $\tau_w$-compact but not $\tau_n$-compact then the family $$\mathscr{C}=\{K : K \text{ is } \tau_n-\text{compact}\} \cup \{K \cup A : K \text{ is } \tau_n-\text{compact}\} $$ satisfies i,ii and iii, but there is no intermediate $\tau_{\mathscr{C}}$ which has $\mathscr{C}$ as its compact sets.