Show that $G = \{x \in \mathbb{R}: 0 < x < 1\}$ is open. 
As i am reading up on introduction to point set topology, i saw this
  example but they did not provide full details. Please help me take a
  look and see if it is correct! Thanks!

We have to make a good gauge and we pick epsilon $\epsilon$ to be either $x$ or $1-x$, whichever is smaller. This means $\epsilon \leq 0.5$. 
Now we need to show that $(x-\epsilon,x+\epsilon)$ is a subset of $G$. Essentially this means that we show for any $u \in (x-\epsilon,x+\epsilon)$, then $u \in G$.
WLOG, we pick $\epsilon = x$, as the other case will be the same.
Now since we know $|u-x| < \epsilon$, we thus have $$|u-x| < \epsilon \Rightarrow -\epsilon < u -x < \epsilon \Rightarrow  0 < u < 2x \leq 1$$
Alternatively, we know from the beginning that $x-\epsilon < u < x+\epsilon $ and we can work from here as well.
This completes the proof as we have shown $u$ is indeed in $G$ for all $u$.

G is the open ball with radius 1/2 centered at 1/2.
