[proofplan] We use the neighbourhood characterization of closure: a point belongs to $\overline{A}$ exactly when every open neighbourhood of that point intersects $A$. First we prove that $A$ is contained in its closure and that the complement of $\overline{A}$ is open. Then we prove minimality by showing that any [closed set](/page/Closed%20Set) containing $A$ also contains every point adherent to $A$. [/proofplan] [step:Use the neighbourhood characterization to show $A \subset \overline{A}$] Let $x \in A$. Let $U \subset X$ be an open neighbourhood of $x$, meaning that $U \in \tau$ and $x \in U$. Since $x \in A$ and $x \in U$, we have $x \in U \cap A$, so $U \cap A \neq \varnothing$. Therefore every open neighbourhood of $x$ intersects $A$, and hence $x \in \overline{A}$. Since $x \in A$ was arbitrary, $A \subset \overline{A}$. [/step] [step:Show the complement of $\overline{A}$ is open] Let $x \in X \setminus \overline{A}$. By the neighbourhood characterization of closure, there exists an [open set](/page/Open%20Set) $U_x \subset X$ such that $U_x \in \tau$, $x \in U_x$, and $U_x \cap A = \varnothing$. We claim that $U_x \subset X \setminus \overline{A}$. Let $y \in U_x$. Since $U_x$ is open and contains $y$, it is an open neighbourhood of $y$. Since $U_x \cap A = \varnothing$, there exists an open neighbourhood of $y$ disjoint from $A$. Therefore $y \notin \overline{A}$. Thus $U_x \subset X \setminus \overline{A}$. For every $x \in X \setminus \overline{A}$, we have found an open set $U_x$ with $x \in U_x \subset X \setminus \overline{A}$. Hence $X \setminus \overline{A}$ is open in $(X,\tau)$. Therefore $\overline{A}$ is closed in $(X,\tau)$. [/step] [step:Prove every closed superset of $A$ contains $\overline{A}$] Let $F \subset X$ be closed in $(X,\tau)$ and suppose $A \subset F$. We prove $\overline{A} \subset F$. Let $x \in \overline{A}$. Assume, for contradiction, that $x \notin F$. Since $F$ is closed, the complement $X \setminus F$ is open. Since $x \notin F$, we have $x \in X \setminus F$, so $X \setminus F$ is an open neighbourhood of $x$. Because $A \subset F$, the open set $X \setminus F$ is disjoint from $A$: \begin{align*} (X \setminus F) \cap A = \varnothing. \end{align*} This gives an open neighbourhood of $x$ that does not intersect $A$, contradicting $x \in \overline{A}$. Therefore $x \in F$. Since $x \in \overline{A}$ was arbitrary, $\overline{A} \subset F$. [/step] [step:Conclude the closure is the smallest closed superset] We have proved that $\overline{A}$ is closed and that $A \subset \overline{A}$. We have also proved that every closed subset $F \subset X$ satisfying $A \subset F$ must satisfy $\overline{A} \subset F$. Therefore $\overline{A}$ is the smallest closed subset of $X$ containing $A$. [/step]