I think singleton sets $\{x\}$ where $x$ is a member of $\mathbb{R}$ are both open and closed.
Singleton sets are open because $\{x\}$ is a subset of itself. There are no points in the neighborhood of $x$.
I want to know singleton sets are closed or not.
As has been noted, the notion of "open" and "closed" is not absolute, but depends on a topology. So in order to answer your question one must first ask what topology you are considering.
A topological space is a pair, $(X,\tau)$, where $X$ is a nonempty set, and $\tau$ is a collection of subsets of $X$ such that:
The elements of $\tau$ are said to be "open" (in $X$, in the topology $\tau$), and a set $C\subseteq X$ is said to be "closed" if and only if $X-C\in\tau$ (that is, if the complement is open).
In $\mathbb{R}$, we can let $\tau$ be the collection of all subsets that are unions of open intervals; equivalently, a set $\mathcal{O}\subseteq\mathbb{R}$ is open if and only if for every $x\in\mathcal{O}$ there exists $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq\mathcal{O}$. You may want to convince yourself that the collection of all such sets satisfies the three conditions above, and hence makes $\mathbb{R}$ a topological space. This topology is what is called the "usual" (or "metric") topology on $\mathbb{R}$.
If you are working inside of $\mathbb{R}$ with this topology, then singletons $\{x\}$ are certainly closed, because their complements are open: given any $a\in \mathbb{R}-\{x\}$, let $\epsilon=|a-x|$. Then $x\notin (a-\epsilon,a+\epsilon)$, so $(a-\epsilon,a+\epsilon)\subseteq \mathbb{R}-\{x\}$; hence $\mathbb{R}-\{x\}$ is open, so $\{x\}$ is closed.
The reason you give for $\{x\}$ to be open does not really make sense. Every set is a subset of itself, so if that argument were valid, every set would always be "open"; but we know this is not the case in every topological space (certainly not in $\mathbb{R}$ with the "usual topology"). So that argument certainly does not work.
So: is $\{x\}$ open in $\mathbb{R}$ in the usual topology? Well, $x\in\{x\}$. Does there exist an $\epsilon\gt 0$ such that $(x-\epsilon,x+\epsilon)\subseteq \{x\}$? If so, then congratulations, you have shown the set is open. If there is no such $\epsilon$, and you prove that, then congratulations, you have shown that $\{x\}$ is not open.
If you are giving $\{x\}$ the subspace topology and asking whether $\{x\}$ is open in $\{x\}$ in this topology, the answer is yes. There is only one possible topology on a one-point set, and it is discrete (and indiscrete). However, if you are considering singletons as subsets of a larger topological space, this will depend on the properties of that space. As Trevor indicates, the condition that points are closed is (equivalent to) the $T_1$ condition, and in particular is true in every metric space, including $\mathbb{R}$. When $\{x\}$ is open in a space $X$, then $x$ is called an isolated point of $X$. If all points are isolated points, then the topology is discrete. In the real numbers, for example, there are no isolated points; every open set is a union of open intervals.
In summary, if you are talking about the usual topology on the real line, then singleton sets are closed but not open.
Every singleton set is closed. It is enough to prove that the complement is open. Consider $\{x\}$ in $\mathbb{R}$. Then $X\setminus \{x\} = (-\infty, x)\cup(x,\infty)$ which is the union of two open sets, hence open. Since the complement of $\{x\}$ is open, $\{x\}$ is closed.
It depends on what topology you are looking at. For $T_1$ spaces, singleton sets are always closed. So for the standard topology on $\mathbb{R}$, singleton sets are always closed.
Assume for a Topological space $(X,\mathcal{T})$ that the singleton sets $\{x\} \subset X$ are closed. Then every punctured set $X/\{x\}$ is open in this topology. If these sets form a base for the topology $\mathcal{T}$ then $\mathcal{T}$ must be the cofinite topology with $U \in \mathcal{T}$ if and only if $|X/U|$ is finite. This is because finite intersections of the open sets will generate every set with a finite complement. Equivalently, finite unions of the closed sets will generate every finite set.
In the space $\mathbb R$,each one-point {$x_0$} set is closed,because every one-point set different from $x_0$ has a neighbourhood not intersecting {$x_0$},so that {$x_0$} is its own closure.
NOTE:This fact is not true for arbitrary topological spaces.
Consider the topology $\mathfrak F$ on the three-point set X={$a,b,c$},where $\mathfrak F=${$\phi$,{$a,b$},{$b,c$},{$b$},{$a,b,c$}}.
Since X\ {$b$}={a,c}$\notin \mathfrak F$ $\implies $ In the topological space (X,$\mathfrak F$),the one-point set {$b$} is not closed,for its complement is not open.
You may just try definition to confirm. Definition of closed set : Let E be a subset of metric space (x,d). E is said to be closed if E contains all its limit points.
Now cheking for limit points of singalton set E={p}, for r>0 , N(p,r) intersection with (E-{p}) is empty equal to phi i.e. set of limit points of {p}= phi so, set {p} has no limit points so clearly {p} contains all its limit points (because phi is subset of {p})
therefor {p} is closed.
