Open Sets Definition I understand that the definition of an open set in a Metric Space and this can transfer over if you're dealing with a Metric Topological Space. However, I'm not sure if there is a standard definition of an open set in a Topological Space. I have read that it may vary under constraints.
I found one definition on Wikipedia:
If ($X$, $D$) is a Topological Space, then a subset $O$ of $X$ is open if and only if $O$ is a neighbourhood of each of its points.
This is how I expressed it mathematically:
for all x $\in$ O there exist an open set U containing x such that U  $\subseteq\ O $

The definition is equivalent to the usual definition in a topological space. When you have a metric space $(X,\rho)$, you can consider the topological space $(X,\tau_\rho)$ where the topology $\tau_\rho$ consists of all sets in $X$ that are unions of open balls $B(x,\varepsilon)$. Observe that
$(1)$ $X$ is a union of open balls, namely $X=\bigcup_{x\in X} B(x,1)$, so $X\in \tau_\rho$.
$(2)$ $\varnothing$ is the empty union of balls, so $\varnothing \in \tau_p$.
$(3)$ The arbitrary union of sets that are union of balls is clearly a union of balls.
$(4)$ If $B(x,\varepsilon_1)$ and $B(y,\varepsilon_2)$ have nonempty intersection, we can always find a point $z$ and a ball $B(z,\varepsilon_3)$ contained in $B(x,\varepsilon_1)\cap B(y,\varepsilon_2)$. This can be used to prove the intersection of two sets that are union of balls is again a union of balls, hence is in  $\tau_\rho$ again.
Thus the tentative topology $\tau_\rho$ that $\rho$ induces is indeed a topology.
When dealing with metric spaces, we say that $O$ is a nbhd of $x$ if it contains an open ball centered at $x$. It can be shown open balls are nbhds of each of its points. Hence, open balls are open using the definition you supply. 
Now, we'd like to prove

PROP A set in a metric space $(X,\rho)$ is open (i.e. it is a nbhd of each of its points) iff it is the union of open balls, that is, iff it is in $\tau_\rho$.

PROOF Suppose the set $S$ is a union of open balls. Pick $x$ in your set. Since $S$ is a union of balls, $x$ must be in some ball in the union, call it $B_1$. But $B_1$ is a nbhd of each of its points, so it contains some open ball containing $x$. This ball will be contained in $S$, so $S$ is open. Conversely, suppose $S$ is a nbhd of each of its points. Then for each $x$ we can find a ball $B(x,\varepsilon_x)$ contained in $S$. Then $S=\bigcup_{x\in S} B(x,\varepsilon_x)$ will be a union of balls $\blacktriangleleft$. 
