This is a question about possible alternative formulations of a compactness.
Suppose $R$ is an open set, and $S$ is a compact subset of $R$. Suppose we have an arbitrary open cover $R = \bigcup_{i\in I} T_i$. Then, as the cover of $R$ also covers $S$ can we, for each such cover construct a finite open cover of $R$ as follows, noting that as $R$ is open and $S$ is closed so $R \setminus S$ is open?
$$R = S \cup (R\setminus S) = \bigcup_{\text{finitely-many } i} T_i \cup (R \setminus S)$$
Ordinarily for compactness one specifies that this is a subcover, so I'm inclined to say that this is not valid for showing that $R$ is in fact compact.
Shifting to the context of modules over a noetherian ring $A$ equipped with the weak topology, if more generally I have that $R$ and $S$ are both clopen $A$-modules with $S \subseteq R$, with $S$ compact, and that for some ideal $I$ of $A$ we have equality of quotients $R/IR = S/IS$. Then is it possible to show that $R$ is compact?
Thanks in advance,
M
To answer the first part of your question, you are correct in noting that this doesn't imply that $R$ is compact since proving compactness requires you to produce a finite subcover of a given cover. In fact, your proof would immediately show that all (non-empty) topological spaces are compact which is of course false :)