A set containing one element is an open set. Why? I asked a question last night about proving that a discrete metric space is both open and closed. Once or twice it was mentioned that a set that contains only one element is open. I'd like to know:
a) is that always true?
b) why?
An explanation that is both a proof and a simple breakdown of it would be most helpful.
 A: In general, it is not the case that single-element sets are open.  For example, in the usual topology on $\Bbb R$, the one induced by the usual metric, single-element sets are not open; open sets are unions of open intervals, and every open set (except $\varnothing$) is infinite.

Say we have a metric space with the so-called "discrete" metric $d$.  Recall that this means that $$d(x,y) = \begin{cases} 0, \text{if $x = y$}\\ 1, \text{if $x\ne y$}\end{cases}$$   
A metric space has a natural topology "induced by" its metric.
The "metric topology" induced by the metric $d$ is the one that has as its basis all "balls" $N_{d,\epsilon}(x)$ where $$N_{d,\epsilon}(x) = \{ p \mid d(x,p) < \epsilon\}.$$  
"Basis" here means that a set is open if and only if it is a union of some of these balls.
The topology induced on $\Bbb R^n$ by its usual metric is exactly the usual topology for $\Bbb R^n$.  But for an unusual metric such as the discrete metric, the situation is different.
Observe that in the discrete metric $d$, $N_{d,1/2}(x) = \{x\}$, because $x$ itself is the only point $p$ such that $d(x,p) < \frac12$.  So each $\{x\}$ is a basis element and is therefore open in the metric topology induced by $d$.
Since any union of open sets is open, and any set at all is a union of sets of the form $\{x\}$, we conclude that any set at all is open in the  topology induced by the discrete metric $d$.
This is why this metric $d$ is called the "discrete" metric: the topology it induces is the discrete topology.
A: I am not sure whether you mean a) and b) in general or just in a discrete metric space so let me answer for both cases
A) for the discrete metric space it is always true by definition. for a general topology it is not
B) consider the ball $B(x,1/2)$. it is a basic open set and contains only one point $x$. if the space is not discrete metric then Real numbers give you an example of why a singleton is not open.
