[proofplan] We prove both directions of the equivalence directly from the definition of closure as the smallest closed set containing $A$. For the forward direction, we show that any point outside $\overline{A}$ has an open neighbourhood disjoint from $A$. For the converse, we show that a point meeting every open neighbourhood of it with $A$ cannot lie in any closed set containing $A$'s complement. [/proofplan] [step:Show that every point of $\overline{A}$ meets $A$ in each of its open neighbourhoods] Suppose $x \in \overline{A}$ and let $U \in \tau$ with $x \in U$. Assume for contradiction that $U \cap A = \varnothing$, so that $A \subset X \setminus U$. Since $U$ is open, the set $X \setminus U$ is closed. The closure $\overline{A}$ is the intersection of all closed sets containing $A$, so $\overline{A} \subset X \setminus U$. This gives $x \in X \setminus U$, contradicting $x \in U$. [guided] We want to show: if $x \in \overline{A}$, then every open set $U$ containing $x$ satisfies $U \cap A \neq \varnothing$. Recall that the closure is defined as \begin{align*} \overline{A} = \bigcap \{ F \subset X : F \text{ is closed and } A \subset F \}. \end{align*} Take any $U \in \tau$ with $x \in U$, and suppose for contradiction that $U \cap A = \varnothing$. Then $A \subset X \setminus U$. The complement $X \setminus U$ is closed (since $U$ is open), and it contains $A$, so it is one of the sets in the intersection defining $\overline{A}$. Therefore $\overline{A} \subset X \setminus U$. Since $x \in \overline{A}$, we get $x \in X \setminus U$, meaning $x \notin U$ — contradicting our assumption that $x \in U$. [/guided] [/step] [step:Show that a point meeting $A$ in every open neighbourhood belongs to $\overline{A}$] Suppose every open set $U \in \tau$ containing $x$ satisfies $U \cap A \neq \varnothing$. Let $F$ be any closed set with $A \subset F$. Then $X \setminus F$ is open. If $x \notin F$, then $x \in X \setminus F$, and $X \setminus F$ is an open neighbourhood of $x$ satisfying $(X \setminus F) \cap A \subset (X \setminus F) \cap F = \varnothing$, contradicting the hypothesis. Hence $x \in F$. Since $F$ was an arbitrary closed set containing $A$, we conclude $x \in \overline{A}$. [guided] Now suppose every open set containing $x$ intersects $A$. We must show $x \in \overline{A}$, i.e., $x$ belongs to every closed set $F$ that contains $A$. Let $F$ be any closed set with $A \subset F$. The complement $X \setminus F$ is open. If $x \notin F$, then $x \in X \setminus F$, which is an open neighbourhood of $x$. By hypothesis, $(X \setminus F) \cap A \neq \varnothing$. But $A \subset F$ forces $A \cap (X \setminus F) = \varnothing$ — a contradiction. Therefore $x \in F$ for every such closed set $F$, and since $\overline{A} = \bigcap \{ F : F \text{ closed}, A \subset F \}$, we have $x \in \overline{A}$. [/guided] [/step]