Why are singletons open in a discrete topology? I was reading through Loring Tu's An Introduction to Manifolds, and I came across this example in the appendix:
For any set $S$, let $\tau$ be the collection of all subsets of $S$. Then $\tau$ is a topology on $S$, called the discrete topology. A singleton set is a set with a single element. The discrete topology can also be characterized as the topology in which every singleton subset $\{p\}$ is open. A topological space having the discrete topology is called a discrete space. The discrete topology is the finest topology on a set.
Right before this example, the author provides a lemma called local criterion for openness, but I am unable to see why a singleton in this situation is open.
EDIT:
Local Criterion for Openness:
Lemma A.2 (Local criterion for openness). Let $S$ be a topological space. A subset $A$ is open in $S$ if and only if for every $p ∈ A$, there is an open set $V$ such that $p ∈ V ⊂ A$.
Proof.
($⇒$) If $A$ is open, we can take $V = A$.
($⇐$) Suppose for every $p ∈ A$ there is an open set $V_p$ such that $p ∈ V_p ⊂ A$. Then $$ A \subset \bigcup_{p \in A} V_p \subset A $$ so that equality $A = \bigcup_{p \in A} V_p$ holds. As a union of open sets, $A$ is open.
 A: In topology the open subsets are nothing more or less then the subsets that we say are open[1].  "Open" by itself doesn't mean anything intrinsic.
Each topology has a list of sets call a "topology" and this list is simply a list of all the subsets that we are going to call open[1].  The discrete topology is one in which the list of all open subset is EVERY subset.  So every set is open.  Why?  Because we said so.  We don't need any other reason[1].
[1]  Okay, the list must obey a few rules.  Any f union of sets on the list must be on the list. And finite intersection of sets must be on the list.  And the set itself and the empty set must be on the list.  But other than that we can select the list to be anything we want.
We can have a "characterisation" of a topology in which we don't list out all the open sets but just a base few, from which by taking unions and intersections we can determine all the other sets on the list.  If we claim that all singletons are open (why?  Because we said so.) then it'd follow that all unions of singletons are open.  Hence all sets are open.
The discrete topology is a topology in which all sets are open.  That is a definition.  You can't prove it because there is nothing to prove.
A: I think you might be overthinking it.  

For any set $S$, let $\tau$ be the collection of all subsets of $S$.

It's standard to use the letter $\tau$ to denote the set of open sets in a given topological space.  In this case, we are told that $\tau$ includes every subset of the set $S$; this is the very definition of the discrete topology.  In particular, singleton sets, i.e. sets of the form $\{x \}$ for any $x \in S$, are themselves subsets of $S$.  So $\{x \} \in \tau$ for every $x \in S$, or in other words, every such set is open.
A: You may wish to verify this topology indeed satisfies the definition of a topology: 


*

*$X \in \tau$ and $\emptyset\in \tau$

*An arbitrary union of elements in $\tau$ is again in $\tau$

*A finite intersection of elements in $\tau$ is again in $\tau$


So we have simply defined all subsets to be in $\tau$ and call them open since we can.
It is perhaps hard to grasp because it is strange to define open sets in this way. It is similar to the first time you see the discrete metric
$$d(x,y) = \begin{cases} 0, \;x \ne y \\
1, \;x=y \end{cases}$$
which doesn't really "seem" like a distance in the intuitive way you know, but satisfies the definition of a metric. 
