So before my book even goes into the theorem that says "A set $E \subset R $ is compact iff E is closed and bounded", they go into examples of an open set, but I'm having trouble understanding their thought process.
Let $E=(0,1]$ and for each positive integer n, let $G_n=(\frac{1}{n},2)$. If $0 <x\le 1$ there is a positive integer $n$ such that $\frac{1}{n}<x$, hence $x \in G_n$ and thus $E \subset \cup_{n=1}^{\infty}G_n$.
Ok, I get that part, basically since $\frac{1}{n}$ goes to $0$ as $n \to \infty$, but never equals $0$, E is contained in the union of $G_n$. However, the next part I don't understand...
"Choose a finite set $n_1,n_2,...,n_r$ of positive integers, then
$\cup_{i=1}^{\infty}G_{n_i}=G_{n_0}$ where $n_0=$max{$n_1,...n_r$}." I don't understand how $G_{n_1}+G_{n_2}+...+G_{n_r}=G_{n_0}$. The largest subcover?
And finally why is $E $ not a subset of $G_{n_0}$
The equality $G_{n_1} \cup G_{n_2} \cup \ldots \cup G_{n_r} = G_{n_0}$ (use union, not addition, these are quite different!) holds because the $G_i$ are increasing: if $i < j$ then ${1 \over j} < {1 \over i}$ so that $G_i = (\frac{1}{i},2) \subset (\frac{1}{j},2) = G_j$.
This means that if we look at any finite set of indices, there is a maximal one among them, and because the larger the index, the larger the set (as we saw), this means that the union of the corresponding $G_i$ is just the one with the largest index, and this is what your text essentially says.
Now, if $G_{n_0}$ is any fixed member of your open cover, then find $x$ with $0 < x < \frac{1}{n_0}$. Then this $x$ is in $(0,1]$ but not in $G_{n_0}$. So one member cannot cover $(0,1]$. So, using what we have seen in the previous paragraph, no finite set of members can cover $(0,1]$, because a finite subset has as its union the one with the largest index, which cannot cover by itself.
This implies that $(0,1]$ is not compact. Compactness of $E$ means every open cover of $E$ has a finite subcover. But here we have some cover, that does not have a finite subcover. And this is exactly what is needed to refute compactness.
Note that by the closed and bounded theorem, which holds in the reals in the usual metric, $[0,1]$ is compact. So just removing a single point (making it non-closed) makes $E$ non-compact, which is the point of the example.