Question: Show that if $(X,\tau)$ is a separable topological space, then $\{x \in X : \{x\} \in \tau\}$ is an enumerable set.

I thought this way, if a topological space is separable, this implies that it has a dense and numerable set, that is, if this set is dense, then its closure is the topological space X, which has numberable points, finite or that has bijeção with the natural numbers.
Does this have anything to do with the elements of the topology being numberable? in particular discrete elements, formed by only one element ${x}$. In fact, how can I state that $\{x\}$ is in the topology?

Answer: 
Let $I$ be the set of isolated points, that is
$I=\{x \in X : \{x\} \in \tau\}$ (all singletons that are open).

Let $D$ be any countable dense set. Then $I$ must be a subset of $D$, hence $I$ must be countable too.

Indeed, suppose the contrary, that $I$ is not a subset of $D$, and pick any $x\in I\setminus D$. Then the set $U=\{x\}$ is open and non-empty, but $U\cap D=\varnothing$, contradiction.

Edit/addition. Just to clarify the definition of separable, since in your question it is not clear whether you think that $D$ is numerable or that $X$ is numerable. A space $X$ (with topology $\tau$) is separable if there is a countable (also called numerable) dense subset $D$ of $X$. That is, $D$ is numerable, and the closure of $D$ equals $X$ (but $X$ need not be numerable). To say that $D$ is dense is the same as to say the $D$ intersects every non-empty open subset of $X$.

Take the real line with the usual topology $\sigma$ and isolate all rational numbers. That is, consider the topology $\tau$ with base $\mathcal B=\sigma\cup\{\{q\}:q \mathrm{\ is\ rational\ }\}$. This provides an example of a separable space $X$ with countably many isolated points, and such that $X$ is uncountable and $\tau$ is uncountable too.
An uncountable space $X$ can have a countable (numerable) dense subset $D$. For example, $X=\Bbb R$ and $D=\Bbb Q.$ And an extreme example: Let $X$ be any non-empty set; let $p\in X;$ let $\tau=\{\emptyset, \{p\}, X\};$ and let $D=\{p\}$.

Let $X^i$ be the set of isolated points of $X$. That is, $X^i=\{p\in X: \{p\}\in \tau \}. $

Let $D$ be dense in $X.$ Then $D\cap U\ne\emptyset$ whenever $\emptyset \ne U\in \tau.$ So if $ p\in X^i$ then $\emptyset \ne \{p\}\in \tau,$ so $\emptyset\ne D\cap \{p\},$ so $p\in D.$ Hence $$X^i\subseteq D.$$ Any subset of a countable set is countable. So if $D$ is countable then $X^i$ is countable.

Appendix. Let $E\subseteq X.$ The definition of $\overline E$ is $\bigcap \Bbb F\_E$ where $\Bbb F\_E=\{F: X\setminus F\in \tau \,\land \, F\supseteq E\}.$ An important fact is that if $U\in \tau$ and if $E\cap U$ is empty then $\overline E\cap U$ is empty. Because $X\setminus U\in \Bbb F\_E,$ so $\overline E=\bigcap \Bbb F\_E\subseteq X$ \ $U.$

Hence, another important fact: If $p\in U\in \tau$ and if $U\cap E$ is empty then $p\not\in\overline E.$ In particular, if $p\in X^i$ and if $p\not\in E$ then $E$ is not dense in $X,$ that is, $\overline E\ne X.$ Because, with $U=\{p\},$ we have $p\in U\in \tau$ and $E\cap U=\emptyset, $ so $\emptyset =\overline E\cap U=\overline E\cap \{p\},$ so $p\not\in\overline E.$