How does one go about proving/disproving that given $(X,d)$ a metric space that a subset $S$ is open.
Given the following definitions:
A set $X$ is open $\iff \forall x \in X, x\in int(X)$ i.e. x is interior point
where an interior point is defined as:
$x$ is a interior point of $A \subset X$ if $\exists r>0, B(z,r)\subset A \land x \in B(z,r) $
I apologise if I am being to vague and/or general
Usually, you can prove a set is open by definition.
For example, in $\mathbb R^2$ with standard metric, the square $S=\{(x,y)| |x|<1, |y|<1\}$ is open. Why? Take any $(x,y)\in S$. Then $|x|<1$, therefore there exists some $\delta_x$ such that $(x-\delta_x, x+\delta_x)\subseteq(-1,1)$. In the same way, there exists some $\delta_y$ such that $(y-\delta_y, y+\delta_y)\subseteq(-1,1)$.
It is then easy to show that for $\delta=\min\{\delta_x,\delta_y\}$, the ball $B((x,y),\delta)$ is a subset of $S$.