
covering
Let $\displaystyle S = (0,1) $ and $\displaystyle \mathcal{O} = \{(1/n,1): n \in \bold{N} \} $. Show that $\displaystyle \mathcal{O} $ is an open cover of $\displaystyle S $.
So choose $\displaystyle x \in (0,1) $. Pick $\displaystyle n_0 \in \bold{N} $ such that $\displaystyle 1/n_0 < x $. This implies that $\displaystyle x \in (1/n_0, 1) \subset \bigcup_{n \in \bold{N}} (1/n,1) = \bigcup_{G \in \mathcal{O}} G $. Is this correct? Could I pick $\displaystyle n_0 $ in a different way?
What if we had a finite subset $\displaystyle \mathcal{O'} \subset \mathcal{O} $?

If you had a finite subset of the collection, then the smallest lefthand endpoint would a number $\displaystyle 0 < t < 1$.
But $\displaystyle \frac{t}{2} \in \left( {0,1} \right)$ and $\displaystyle \frac{t}{2} < t$.
Therefore, the subcollection does not cover $\displaystyle \left( {0,1} \right)$.

The statement that set is compact means every open covering has a finite subcover.
Having one open cover which has a finite subcover does not imply that the set is compact.
Consider: $\displaystyle \left( {  1,.000009} \right) \cup \left[ {\bigcup\limits_{n \in N} {\left( { n^{  1} ,1} \right)} } \right]$
That is clearly an open covering of $\displaystyle \left( {0,1} \right)$ and the collection contains a finite subcover.
But $\displaystyle \left( {0,1} \right)$ is still not compact.