Need help on understanding Heine-Borel property. Heine-Borel Theorem states that if a set has an open cover and if we can find a finite subcover from that open cover that covers the set, the set would be compact. 
I got this question while I was trying to prove that Heine-Borel property will imply that the set is closed.
Is there any restriction on open sets that are the contained in the open cover? If not, can't I just make one of the open sets to be a set of real numbers? Then, the set would always have a finite subcover, that is, the set of all real numbers that covers the set. Hence, it can cover $(0,1)$ or any other open intervals but still $(0,1)$ is not closed.
 A: That is not what the Heine-Borel Theorem states. The Heine-Borel Theorem states that if $S\subseteq \mathbb{R}^n$ (with the usual topology/metric), then $S$ is compact if and only if $S$ is closed and bounded. 
What you stated is more like the definition of compact: $S$ is compact provided that for every open cover of $S$, there is a finite subcover. That "every" is extremely important. Sure, $\{(0,1)\}$ is an open cover of $(0,1)$ and it has a finite subcover, but there exist open covers of $(1,2)$ with no finite subcover. Thus $(0,1)$ is not compact.
A: The Heine Borel Theorem states that in $\mathbb{R}^n$, a set is compact iff it is closed and bounded. Perhaps in your class you initially defined compactness as being closed and bounded, but in general a set $K$ is compact if every open cover has a finite sub-cover. 
An open cover $\mathcal{U}$ for as set $K$ is a collected of sets that are open in $\mathbb{R}^n$ such that $$K\subset\bigcup_{U\in\mathcal{U}}U.$$ So, the sets simply must be open. I think the key point you are missing here is that all open covers have to have a finite subcover, not just some open cover. Otherwise, every subset of $\mathbb{R}^n$ would be compact, because $\mathcal{U}=\{\mathbb{R}^n\}$ is an open cover for every subset of $\mathbb{R}^n$. 
To see that every compact set is closed, suppose $K$ is compact and $p\not\in K$. For each $x\in K$, let $U_x$ be an open set containing $x$ and $V_x$ be an open set containing $p$ such that $U_x\cap V_x=\emptyset$. Then $\mathcal{U}=\{U_x:x\in K\}$ is an open cover of $K$. Thus, we can extract a finite subcover $\{U_{x_1},\dots, U_{x_n}\}$. Note that $$\left(\bigcap_{i=1}^n V_{x_i}\right)\cap\left(\bigcup_{i=1}^n U_{x_i}\right)=\emptyset$$ by our construction. Since $K\subset \bigcup_{i=1}^n U_{x_i}$, it follows that $V=\bigcap_{i=1}^n V_{x_i}$ is an open set containing $p$ that contains no element of $K$. Thus, $p$ cannot be a limit point of $K$. It follows that $K$ contains all of its limit points, and is closed.
Note that the point where we used a finite subcover is so that we could take a finite intersection of open sets to ensure $V$ was open. 
A: "Heine-Borel Theorem states that if a set has an open cover and if we can find a finite subcover from that open cover that covers the set, the set would be compact."
1) that's not the theorem that states that but the definition of compact and 
2) VERY IMPORTANT  It's not that if you can fine one open cover with a finite subcover.  It's that EVERY open cover must have a finite subcover.
So, yes, $\{\mathbb R\}$ is an open cover of $(0,1)$ and $\{\mathbb R\}$ is itself finite.  That is true$*$.  
But $\{\mathbb R\}$ is only one open cover.  There are others.  And all the others must also have finite subcovers for $(0,1)$ to be compact.
$\{(\frac 1n, 1)| n \in \mathbb N\}$ is a different open cover.  It does not have a finite subcover.  So it is a counter example that proves that not every open cover of $(0,1)$ has a finite subcover.
So $(0,1)$ is not compact.  And by the Heine-Borel Theorem is not closed and bounded.
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$*$[  $\{\mathbb R\}$ is an open cover for every subset of $\mathbb R$ which means every set is compact which means every set is closed and bounded which ... not only is obviously not true, would make the word "compact" utterly useless.  Also "closed" and "bounded" would be meaningless.]
A: It has to work for any open cover rather than just a particular open cover. 
For exampl, consider the open cover $\{\left( \frac1{N}, 1 \right):N \geq 1\}$, finitely many of them will not cover $(0,1)$.
