Show that the closure of a set A is the smallest closed set containing A. I need to prove that $\bar A$ (the closure of set A) is the smallest closed set containing A. We have already proved that it is a closed set, so now I just need to show that it is the smallest one.
I have written a preliminary proof that I don't think is particularly rigorous, and would be grateful if someone could give me some pointers. We haven't covered anything regarding metric spaces or anything else in topology, so I had trouble understanding other solutions posted on the site.

Let $A$ be a non-empty set, and $\bar A$ the closure of $A$ (the union of $A$ and all of its limit points). Let $B$ be a closed set with $A \subset B \subset \bar A$ and $B \neq \bar A$.
Since $A \subset B$ and $B \subset \bar A$, $B$ consists of all the elements of $A$ and some (but not all) of its limit points. However, this means that there are sequences contained entirely within $B$ whose limit points are not elements of $B$. Thus, $B$ is not closed, posing a contradiction to the original statement.

 A: Say $\bar A =  \cap \{ F \supseteq A : F \text{ closed}\}$.
We know $A$ is a closed.
If $x \in \bar A$ then there is sequence $(x_n)$ with ${x_n} \to x$.
Then ${x_n} \in F$ but also $F$ closed set.
So $x = \mathop {\lim }\limits_{n \to \infty } {x_n} \in F$.
That is $\bar A \subseteq F$.
So $\bar A \subseteq  \cap \{ F \supseteq A:F \text{ closed}\}$
A: Yes, your proof is quite correct. Maybe one suggestion would be to make it clear immediately that you are going for a contradiction.

... Let Suppose that there is some closed set be $B$ be a closed set with such that $A \subset B \subset \bar A$ and $B \neq \bar A$. ...

A: I think it is clearer without a contradiction argument:

Let $B$ be a closed set satisfying $A\subset B$. Let $a$ be a limit point of $A$, so that we can find a sequence $(a_{n})_{n}\subset A$ with $a_{n}\rightarrow a$. Since this sequence is also contained in $B$ and $B$ is closed, it follows that $a \in B$. Because $a$ was an arbitrary limit point, $B$ contains the closure of $A$.

A: If $x$ is a limit point of $A$ and $A\subset B$, where $B$ is closed, then $x$ is a limit point of $B.$ Since $B$ is closed $x\in B$. This means that the closure of a set $A$ is the smallest.
A: Let $\overline{A}$ be the closure of a set $A\ne\emptyset$. Suppose $\overline{A}$ is not the smallest closed set containing $A$. Then, $\exists$ some closed set, $B$ such that
$$A\subseteq B\subset \overline{A}.$$
Since $A\subseteq B$ and $B\subset\overline{A},~B$ consists of all the elements of $A$ and some (but not all) of its limit points. (If it contained all its limit points, then $B\supseteq \overline{A}$, contradicting our assumption.) There exists a limit point, $x$ of $A$ not in $B$. Since $x$ is a limit point of $A$ and $A\subseteq B$, we have
$$\forall\epsilon>0~\exists y\ne x~\backepsilon y\in A\cap B(x,\epsilon)\Rightarrow\forall\epsilon>0~\exists y\ne x~\backepsilon y\in B\cap B(x,\epsilon)$$
i.e., $x$ is also a limit point of $B$. $B$ is a closed set and hence, contains all its limit points. We get that
$$x\in B,\mathrm{~~which~is~a~contradiction.}$$
Therefore, our assumption is wrong and $\nexists$ a closed set $B$ such that
$$A\subseteq B\subset \overline{A}.$$
$\therefore\overline{A}$ is the smallest closed set containing $A$.
