An open cover of $\{x:x>0\}$ with no finite subcover Two parts. Second one doesn't really warrant its own thread but I want to confirm it.
First: Find an open cover of $ \{ x: x>0\} $ with no finite subcover. I don't have any answer to this one.
Second: Find an open subcover of $(1,2)$ with no finite subcover. I believe an answer to this is $\bigcup_{n=1}^\infty \left( 1 +\frac{1}{n}, 2-\frac{1}{n} \right)$, but I'm not sure. 
 A: How about $\bigcup_{n=1}^{\infty}(0,n)$ for the first one. Your cover for the second one works.
A: For the second one. Your answer is correct. A cover $\mathcal{U}$ of the set $(1,2)$ is a collection of open sets whose union contains $(1,2)$. The following equality is clear (we start the union at two to avoid the empty case).
$$ \bigcup_{n=2}^\infty \left(1+\frac{1}{n},2-\frac{1}{n}\right)=(1,2) $$
So the family $\mathcal{U}$ of open sets of the form $ \left(1+\frac{1}{n},2-\frac{1}{n}\right)$ for $n\geq 2$ gives a cover of $(1,2)$. Now suppose you have a finite subcover. That is, a subfamily $\mathcal{V}$ made of members of $\mathcal{U}$ whose union also contains $(1,2)$. Since the set is finite and all of the sets are of the form $(1+1/n,2-2/n)$ then you can find one of this sets that contains all of the others (the one with the biggest value of $n$), say $(1+1/n^*,2-2/n^*)$. But then the union of all the elements of $\mathcal{V}$ is  $(1+1/n^*,2-2/n^*)$ which does not contain the interval $(1,2)$ and gives a contradiction.
For the first one: Use the same idea, how about sets of the form $(1/n,\infty)$. They are open and the union of all of this set is the set $\{x \mid x>0\}$.
The proofs you are for asking are related to compactness. You should have a look at the concept. Note that $(0,1)$ and $(0,\infty)$ are very similar. But $[0,1]$ is not. You can infinitely approach $(0,1)$ from the inside with sets of the form $(0,1-1/n)$, the same happens when you approach $(0,\infty)$ with sets of the form  $(0,n)$. But if you ever reach $[0,1]$ with sets of the form $[0,1/n]$ the moment you do it you can stop and have a finite subcover. This is in fact Borel's original insight that is reproduced in Heine-Borel's theorem that characterizes the compact sets is $\mathbb{R}$ as those that are closed and bounded.
