Must the index for the sets in a finite open subcover $\mathscr{G}$ be arbitrary or can it be a specific number? I have a question regarding a proof in my text proving that the set $S = (1,2)$ is not compact without using the Heine-Borel theorem.
Note: This is a real analysis related question.
Here is the definition of compactness used in the text (the discussion of compact sets is prior to the section on topology, so the definition of compactness presented here makes no reference topological spaces).
The definition of compactness in my text is that a set $S$ is said to be compact if whenever it is contained in the union of a family $\mathscr{F}$ of open sets, it contained the union of some finite number of the sets in $\mathscr{F}$. Equivalently, the set $S$ is compact iff every open cover of $S$ contains a finite subcover.
A suitable cover of $S$ can be constructed from the family $\mathscr{F}$ of sets defined as $\mathscr{F} = \{A_n \, : \, n \in \mathbb{N} \}$ where  $A_n = \left(\frac{1}{n}, 3 \right)$. But here is the part where I have my question concerning the following excerpt from the text

However, if $\mathscr{G} = \{A_{n_1} \dotsc A_{n_k}\}$ is any finite subfamily of $\mathscr{F}$, and if $m = \mathrm{max}\{n_1, \dotsc, n_k\}$, then
$$A_{n_1} \cup \cdots \cup A_{n_k} = A_m = \left(\frac{1}{m},3\right)$$
It follows that the finite subfamily $\mathscr{G}$ is not an open cover of $(0,2)$.

I have no problem understanding the proof in the above quote, but must every demonstration showing that a set is not compact have an aribtary but fixed index for the finite subcover $\mathscr{G}$? That is, must we have $n_k$ as an arbitrary but fixed index for $\mathscr{G} = \{A_{n_1} \dotsc A_{n_k}\}$, or is it just as fine to choose a particular number for $k$, say 5, so that $\mathscr{G} = \{A_{n_1} \dotsc A_{n_5}\}$ where $\mathscr{G}$ is sufficient to demonstrate that $(1,2)$ is not compact?
 A: You can turn your question around by contraposition to get the following 
claim:

If $X$ is compact, then every open cover has a subcover of size $<n$.

It is clear that this statement is false for each $n$ - for example, the discrete topology on a set of size $n$ is compact, but the open cover $\{\{x\}:x\in X\}$ has size $n$ and has no proper subcovers. (You also don't need the general topological definition here - even in $\Bbb R$ you could just consider the set $\{1,\dots,n\}$ which is compact but has the open cover consisting of $(k-1/2,k+1/2)$ for $1\le k\le n$, which has the same properties.)
In other words, given a compact set you know that any open cover has a finite subcover, but you do not have any control on how large the finite subcover is, so a proof that a certain set is not compact must allow the purported finite subcover to be arbitrarily large.

It is possible to consider a definition which allows you to make this assumption, for example if I say that a set is $n$-compact if every open cover has a subcover of size at most $n$, but this definition will not have too many redeeming features. Edit: Turns out this already exists, with the given name; it is very similar to the Lindelöf number which is the smallest $\alpha$ such that $X$ is $\alpha$-compact. The usual $\alpha$ of interest, though, are $\aleph_0$ (compact) and $\aleph_1$ (Lindelöf), not finite cardinalities. The union of $n$-compact sets is not $n$-compact - instead we have that an $m$-compact and $n$-compact set give an $m+n$-compact set. Also, in most usual settings the definition trivializes to the same as just being finite of size $\le n$:

If $X$ is $T_1$ and $n$-compact, then $|X|\le n$.

Proof: Let $Y$ be a subset of $X$ of size $n+1$, and let $U_x=X\setminus (Y\setminus x)$ for each $x\in Y$. Then each $U_x$ is open since $Y\setminus x$ is finite and hence closed, but any open subcover must contain $x$ for each $x\in Y$, and hence must also contain $U_x$ (because the other $U_y\ne U_x$ all miss $x$). Thus $(U_x)_{x\in Y}$ has no subcover of size $\le n$.
