Closure of a subset Show that the closure $\overline E$ of a subset $E \subset \mathbb R^n$ is the smallest closed set containing $E$.
I am new to topology so I was wondering how to establish a solid proof of this statement that seems basic.
Here is my attempt:
First we can show that $\overline E$ is a closed set containing $E$ :

*

*$\overline E = E \cup \partial E $ so it surely contains $E$.

*$\overline E^C = E^C \cap (\partial E)^C$. I need to show that $\forall \vec{x} \in \overline E^C, \exists \delta_x > 0 \text{ such that } B(\vec{x},\delta_x) \subset \overline E^C$ but I struggle to show it properly.

*Then I need to show that $\overline E$ is the smallest set with the previous conditions. I wanted to do it by contradiction by supposing that $\exists F \subset \overline E$ (so smaller than $\overline E$) and then find a contradiction but it seems too strange to prove.

 A: For your attempt toward contradiction, you assume that $F$ contains $E$, is closed, and is strictly contained in $\overline{E}$. If such a set could exist, then $\overline{E}$ would not be the smallest closed set containing $E$. There are several equivalent definitions of the closure, this being one of them, but I will proceed with the fact that $\overline{E}$ is the set $E$ together with all its limit points.
Assuming $E \subset F \subset \overline{E}$ with $F$ closed, then every sequence in $E$ is also a sequence in $F$. Using the fact that closed sets contain their limit points and the aforementioned definition of the closure, can you arrive at a contradiction from here?
A: First show that $\overline{E} = E \cup \partial E$ is closed: so take any $x \notin \overline{E}$. So in particular $x \notin E$ and $x \notin \partial E$. Consider balls $B(x,r)$ around $x$: they will always contain a point of $E^\complement$, namely $x$, so because we know that $x \notin \partial E$, which is defined as the set of all $y$ such that all balls around $y$ contain points from $E$ and also from $E^\complement$, there must be some ball $B(x,r_0)$ that does not contain points from $E$. But the last thing just says that $B(x,r_0) \subseteq E^\complement$, and it also implies $B(x,r_0) \subseteq \partial E^\complement$ (any $y$ in $B(x,r_0)$ has a small ball around it that still sits inside $B(x,r_0)$ so misses $E$ and so $y$ cannot be in $\partial E$ by definition), and it follows that $B(x,r_0) \subseteq \overline{E}^\complement$, and as $x$ was arbitrary there, $\overline{E}^\complement$ is open and $\overline{E}$ is indeed closed.
Now "if $C$ is closed and $E \subseteq C$ then $ \overline{E} \subseteq C$ is the final fact to be shown, which demonstrates the minimality.
So suppose $x \in \overline{E}$ but $x \notin C$. As $C$ is closed and the complement of $C$ is open, we can find some $r_0 >0$ such that $B(x, r_0) \cap C = \emptyset$. It follows that $x \notin E$ (as $ \subseteq C$) and $x \notin \partial E$ (as witnessed directly by this $r_0$) so $x \notin \overline{E}$, contradiction. So $x \in C$ and we're done.
