Proof that the closure is closed

Proof

1. Let $(X, \tau)$ be a topological space and let $S \subseteq X$.
2. Let $\mathcal{F}$ be the family of all closed sets $F$ of $X$ with $S \subseteq F$. By the definition of the closure, $\overline{S}$ is the intersection of the members of $\mathcal{F}$.
3. The family $\mathcal{F}$ is nonempty. Indeed $X \setminus X = \varnothing$ is in $\tau$, so $X$ is closed, and $S \subseteq X$.
4. We claim that $X \setminus \overline{S}$ is the union of the complements $X \setminus F$ for $F \in \mathcal{F}$.
5. Let $x \in X$. Then $x \notin \overline{S}$ if and only if $x \notin F$ for some $F \in \mathcal{F}$, that is, $x \in X \setminus F$ for some $F \in \mathcal{F}$. This proves the claim.
6. For each $F \in \mathcal{F}$, the set $X \setminus F$ is open, because $F$ is closed.
7. By the second axiom of a topology, a union of members of $\tau$ is in $\tau$. By steps 4 and 6, $X \setminus \overline{S}$ is open.
8. By the definition of a closed set, $\overline{S}$ is closed in $X$.