Show that $(0, 1)$ fails Heine-Borel. So I think I understand the Heine-Borel property, but I want to make sure I've wrapped my head around it. With the help of @C Squared I was able to come up with the following, but I want to make sure it's okay. Thanks!
Let $I_k = (\frac{1}{k}, 2)$. Observe that $\bigcup_{k=1}^\infty I_k$ covers $(0, 1)$. Let $x\in (0, 1)$. It follows by the achimedean principle that there exists some natural number $k$ such that $\frac{1}{k} < x$. We may now conclude,
\begin{equation*}
0 < \frac{1}{k} < x < 1 < 2
\end{equation*}
\begin{equation*}
x \in (\frac{1}{k}, 2), k \in \mathbb{N}^+.
\end{equation*}
so,
\begin{equation*}
x\in\bigcup_{k=1}^\infty (\frac{1}{k}, 2).
\end{equation*}
therefore,
\begin{equation*}
(0, 1) \subset (1, 2) \cup (\frac{1}{2}, 2) \cup  (\frac{1}{3}, 2) \cup \dots = \bigcup_{k=1}^\infty I_k
\end{equation*}
Now, Heine-Borel insists that there exists a finite subcover for every cover of $(0, 1)$ that also covers $(0, 1)$. So to show that $(0,1)$ is not compact we must show that the finite subcovers of $\bigcup_{k=1}^\infty I_k$ do not cover $(0,1)$.
Suppose a finite subcover ($\bigcup_{k=1}^n I_k$) of $\bigcup_{k=1}^\infty I_k$. Then we have,
\begin{equation*}
(0, 1) \not\subset \bigcup_{k=1}^n (\frac{1}{k}, 2),
\end{equation*}
because $\frac{1}{n+1} \in (0, 1)$ but $\frac{1}{n+1} \not\in (1, 2) \cup \dots \cup (\frac{1}{n}, 2)$ (by the definition of this set).
There exists a finite subcover that does not cover $(0, 1)$, so the set fails Heine-Borel and is not compact.
 A: Using the Heine-Borel theorem, you need only show that $(0,1)$ is not closed. Since $0,1$ are limit points of $(0,1)$, and $0,1\not\in[0,1]$, then $(0,1)$ is not closed, therefore it cannot be compact.
Alternatively, if you suppose that $(0,1)$ is compact, then consider the cover $$\mathcal{U}=\bigcup_{k=1}^{\infty}(\frac{1}{k},2)$$ By assumption, $\mathcal{U}$ admits a finite sub-cover, $$\mathcal{U}'=\bigcup_{k=1}^n(\frac{1}{k},2)$$ for some $n\in\mathbb{N}$. Now, notice that $(\frac{1}{n},2)=\mathcal{U}'$.
We claim that $(0,1)$ is not a subset of $(\frac{1}{n},2)$.
Fix $n\in\mathbb{N}$. Since $\frac{1}{n+1}\in (0,1)$ and $\frac{1}{n+1}\not\in(\frac{1}{n},2)$, then $(0,1)$ is not a subset of $(\frac{1}{n},2)$, so $\mathcal{U}'$ is not a cover of $(0,1)$. Since $n$ was arbitrary, then no finite sub-cover exists.
Because we have found a cover of $(0,1)$, namely $\mathcal{U}$, that does not admit a finite sub-cover, then $(0,1)$ cannot be compact.
A: As a subset of $\mathbb R$, $(0,1)$ is not compact.  That is because $\mathbb R$ has the Heine-Borel property and $(0,1)$ is not closed and bounded.  And you successfully gave an example of an open cover with no finite subcover.
But that is not what the question is asking.  The question is asking to think of $(0,1)$ as a space in and of itself, does not have the Heine-Borel property (which means that in $(0,1)$ as a space that the Heine-Borel Theorem is FALSE).
And to do that we consider the the set $(\frac 0, \frac 12]$.  This set is CLOSED in $(0,1)$ because it contains all its limit points (note $0$ does not exist in $(0,1)$ so $0$ is not a limit point of $(\frac 0, \frac 12]$).  And the set is bounded.
But it is not compact. Let $I_n = (\frac 1{n+2}, 1 - \frac 1{n+2})$.  $\{I_n\}$ is an open cover of $(\frac 0, \frac 12]$.  (For any $x \in (0, \frac 12]$ there is a $k\ge 3$ so that $\frac 1k < x$ so $x \in I_{k-2}$).  But it has no finite subcover as for any finite $\{I_n\}$ there is a $m= \min n$ and so if $0< x < \frac 1{m+2}$ we have $x$ not being covered.
So $(0,1)$ doesn't have the Heine-Borel Property because in $(0,1)$ the Heine Borel theorem is FALSE.
