Answer by Arturo Magidin for $x\in \overline A \iff$ - how to derive definition of closure

Here are the definitions you are say you are using, once we dig through the comments (please put them in the post, this time and next!)

1. For $\epsilon\gt0$ and a point $x$, $V_{\epsilon}(x) = \{y\mid d(x,y)\lt \epsilon\}$.

2. For a set $A$, the derived set of $A$ is $$A'=\bigl\{x\;\bigm|\; \forall\epsilon\gt0\ \bigl( (V_{\epsilon}(x)\cap A)\setminus\{x\}\neq\varnothing\bigr)\bigr\}.$$

3. For a set $A$, the closure of $A$ is defined to be $\overline{A}=A\cup A'$.

For simplicity, let$$F=\{x\mid \forall\epsilon\gt0 \ (V_{\epsilon}(x)\cap A\neq\varnothing)\}.$$We want to prove that $F=\overline{A}$.

To prove that $\overline{A}\subseteq F$, note that if $x\in A'$, then by definition we have that for all $\epsilon\gt0$, $V_{\epsilon}(x)\cap A\neq\varnothing$, so $x\in F$. Thus, $A'\subseteq F$. And if $x\in A$, then for all $\epsilon\gt 0$, $x\in V_{\epsilon}(x)\cap A$, so $x\in F$. Thus, $A\subseteq F$ also holds. Thus, we have$$\overline{A} = A\cup A'\subseteq F.$$

Conversely, let $x\in F$; to prove that $x\in\overline{A}$, we need to show that $x\in A$ or that $x\in A'$. If $x\in A$, then we are done. So assume that $x\in F$ and $x\notin A$. Let $\epsilon\gt 0$. We want to show that $(V_{\epsilon}(x)\cap A)\setminus\{x\}\neq\varnothing$.

Because $x\in F$, we know that $V_{\epsilon}(x)\cap A\neq\varnothing$. And because we are assuming that $x\notin A$, then it follows that $x\notin V_{\epsilon}(x)\cap A$. Thus, $(V_{\epsilon}(x)\cap A)\setminus\{x\} = V_{\epsilon}(x)\cap A \neq\varnothing$. We have thus shown that the intersection of $V_{\epsilon}(x)$ with $A$ contains points other than $x$. As this holds for every $\epsilon\gt 0$, it follows that $x\in A'$.

Thus, we have shown that if $x\in F$, then either $x\in A$ or else $x\in A'$. Thus, $F\subseteq A\cup A'= \overline{A}$.

This proves that $F=\overline{A}$, as desired.