To prove the Heine-Borel theorem you need to show that a compact set is both closed and bounded. There is a proof of the theorem in the book The Elements of Real Analysis by Bartle. In the proof to show that a compact set K is closed, a specific open cover is used: $$ G_{m} =\left \{ y \, \epsilon \, \mathbb{R}^{n}: ||y-x|| > 1/m \: \forall m \, \epsilon \, \mathbb{N} \right \}$$ where $$ x \, \epsilon \, \complement_{K} $$
I'm curious as to why we are allowed to use a specific open cover to prove a compact set is closed. My thought was that a general open cover should be used because compactness of a set is a property which is related to every open cover on that set.
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$\begingroup$The open cover that you mentioned are used to prove that if it is a compact set, then it is closed and bounded. Hence a particular open cover can be used.
The $G_{\alpha}$ in the proof in the converse direction should be viewed as a general open cover.
$\endgroup$ $\begingroup$Bartle considered these $G_m$'s to show $C(K)$ is open, as he wanted to show every $x\in C(K)$ lies in an open set which doesn't intersect $K$.
$\endgroup$ $\begingroup$Given a union of open intervals $\bigcup_i (A_i,B_i)$ such that $\bigcup_i (A_i,B_i) \supset [A,B]$ we can take finite number of them $\bigcup_{i\le N} (A_i,B_i)$ such that $\bigcup_{i\le N} (A_i,B_i) \supset [A,B]$
Proof:
Step 1: Cover $A$ with an open interval $(A_i,B_i)$
Step 2: Choose an open interval $(A_{i_1},B_{i_1})$ that intersect with $(A_i,B_i)$ to the left for which
case 1: Length $L=\mu((A_{i_1},B_{i_1})\bigcap (A_i,B_i)^c)$ is maximum if the maximum exists. If $B$ has not been covered repeat step 2 with $(A_{i_1},B_{i_1})$ and $(A_i,B_i)$ replaced with $(A_{i_2},B_{i_2})$ and $(A_{i_1},B_{i_1})$ respectively.
case 2: if the maximum of $L$ doesn't exist that means there is a subsequence of intervals that intersect with $(A_i,B_i)$ for which the subsequence {$(A_{i_m},B_{i_m})$}for which {$B_{i_m}$} is increasing and the limit is $k$, since $k$ is also covered by a certain interval $(A_{i_{2}},B_{i_{2}})$ we can find an interval $(A_{i_1},B_{i_1})$ from the subsequence {$(A_{i_m},B_{i_m})$} that intersects both $(A_{i_{2}},B_{i_{2}})$ and $(A_i,B_i)$. If $B$ has not been covered Repeat step 2 with $(A_{i_1},B_{i_1})$ and $(A_i,B_i)$ replaced with $(A_{i_3},B_{i_3})$ and $(A_{i_2},B_{i_2})$ respectively.
It must be that $\lim_{n \to\infty} B_{i_{n}} > B$ , if not that would have implied that $B$ is not covered in any open interval. Which means there is an $N$ such that $B_{N}>B $ and $\bigcup_{i\le N} (A_i,B_i) \supset [A,B]$
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