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 $u$.
 A: Your argument is valid. You didn't need the WLOG though; what you wanted to prove was that $u$ is in $(0,1)$. You know that $|u-x|<\epsilon$ therefore, $-\epsilon<u-x<\epsilon$. Adding $x$ yields $$x-\epsilon<u<x+\epsilon\quad (*)$$. Now, $\epsilon=\min\{x,1-x\}$ means that $x+\epsilon\leq x+1-x=1$ and $x-\epsilon\geq x-x=0$. Replacing this to $*$ brings you to
$$0<u<1$$
as desired.
A: Well you can circumvent all of this if you know that the order topology $\mathcal{T}$ on $\mathbb{R}$ has basis $\mathcal{B} = \{(a,b) \ | \ a, b \in \mathbb{R}\}$.
And it it a well-known (and easy to prove) theorem that the order topology on $\mathbb{R}$ and the metric topology (which is the topology you are using) on $\mathbb{R}$ are equivalent.
Therefore any open interval of the form $(a, b)$ is open in $\mathbb{R}$ endowed with either the order or metric topology. Hence $G = \{x \in \mathbb{R} | 0 < x< 1\}$ is open in $\mathbb{R}$ with either of these topologies.

Sidenote: If you're working with $\mathbb{R}$ or any arbitrary set $X$, please mention the topology endowed on the set $X$, after all a topological space is an ordered pair $(X, \mathcal{K})$ where $\mathcal{K}$ is the topology on $X$. The reason I mention this is that $G$ is not universally open in every topology endowed on $\mathbb{R}$, for example $G$ is neither open nor closed in the trivial topology on $\mathbb{R}$, which is given by $T' = \{\mathbb{R}, \emptyset\}$
A: G is the open ball with radius 1/2 centered at 1/2.
