Basic properties of closed sets

Theorem
Let $$X$$ be a topological space and $$A$$ a subset of $$X$$. Then
 * 1) $$A\subseteq \overline A$$
 * 2) If $$(x_n)$$ is a convergent sequence of points in $$A$$, its limit is an element of $$\overline A$$.
 * 3) If $$X$$ satifies the first countability axiom and $$x\in \overline A$$, then there exists a convergent sequence $$(x_n)$$ of points in $$A$$ such that $$x_n\to x$$.

Proof
1. The definition of adherent point implies immediately that all points of $$A$$ are adherent points for $$A$$.

2. Let $$(x_n)$$ be a convergent sequence of points in $$A$$ and $$x$$ its limit, and let $$V\in \mathcal V(x)$$. By definition of limit there exists $$m\in \mathbb N$$ such that for all $$n>m$$, $$x_n\in V$$. But, by hypothesis, for all $$n>m$$, $$x_n\in A$$, hence $$V\cap A\not =\emptyset$$, meaning that $$x$$ is an adherent point for $$A$$.

3. Let $$x\in \overline A$$ and consider a countable fundamental system of neighbourhoods of $$x$$, $$\{B_n^'\}_{n\in \mathbb N}$$ (which exists by definition). Construct a chain (ordered by inclusion) $$\{B_n\}_{n\in \mathbb N}$$ by setting for all $$n\in \mathbb N$$. Note that this chain is still a fundamental system. Now for each $$n$$, let $$x_n\in B_n$$. Let us prove that $$x_n\to x$$. Suppose $$V\in \mathcal V(x)$$. Then (by definition) there exists $$m\in \mathbb N$$ such that $$B_m\subseteq V$$. But since $$\{B_n\}_{n\in \mathbb N}$$ is a chain, we have and hence $$\underset{p>m}{\forall}\; x_p\in V$$ which ends the proof.
 * $$B_n=\bigcap_{m\leq n} B_m^'$$
 * $$\underset{p>m}{\forall}\; B_p\subseteq V$$,