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Let $(X,\tau)$ be topological space and $A \subset X$. I'm working with this definition for closure: $\overline{A} := A \cup \partial(A)$.

Proof: Note $\overline{A}^\complement = A^\complement \cap \partial(A)^\complement$. If the complement is empty, then we are done since it implies the closure closed. Thus, assume $\overline{A}^\complement \neq \varnothing$. So, for all $x \in \overline{A}^\complement$, we have $x \notin A$ and there exists $u_x \in \tau$ such that $x \in u_x$ and either $u_x \cap A = \varnothing$ or $x \in u_x \cap A^\complement = \varnothing$. Yet, $x \notin A$, so $u_x \cap A = \varnothing$ and $u_x \subset A^\complement$. Hence, for every $x \in \overline{A}^\complement$ there exists $u_x$ where $$\bigcup_{x \in \overline{A}^\complement} u_x = \overline{A}^\complement.$$ Since $\overline{A}^\complement$ is the union of open-neighborhoods of $X$, then it is also open. Therefore, $\overline{A}$ is closed.

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  • $\begingroup$ From $x\notin A$ you conclude that $u_x\cap A=\varnothing$ without any argumentation. $\endgroup$ – drhab Feb 9 '18 at 16:12
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I missed something in your answer. See my comment.

Let $x\in\overline A^{\complement}=A^{\complement}\cap(\partial A)^{\complement}$.

If for every $U\in\tau$ that contains $x$ it is true that $U\cap A\neq\varnothing$ then the conclusion $x\in\partial A$ is justified. This because in that case we have $U\cap A\neq\varnothing$ and $U\cap A^{\complement}\neq\varnothing$ for every $U\in\tau$ with $x\in U$.

So the fact that $x\in(\partial A)^{\complement}$ tells us that there must be an open set $u_x$ that contains $x$ and satisfies $u_x\cap A=\varnothing$. Elements in $u_x$ will not be elements of $\partial A$ because $u_x$ serves as a neighborhood of these elements that has empty intersection with $A$. So actually we have $u_x\subseteq A^{\complement}\cap(\partial A)^{\complement}=\overline A^{\complement}$

That leads to: $$\overline A^{\complement}=\bigcup_{x\in\overline A^{\complement}}u_x\in\tau$$

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