Prove that $\bar S = S \cup [\text{Accumulation points of S}]$ Let $S \subseteq E$, where $E$ is a metric space. Then

$x\in S$ is an accumulation point of $S$ iff there is a sequence $x_n \in S$ such that $\lim x_n = x$ and $a_n \not = a_m$ for $n \not = m$. (Real Analysis, T. Terzioğlu, p24)



And we already know that 

$x\in \bar S$ iff there exists an sequence $x_n \in S$ s.t $\lim x_n =
x$.

And I want to prove the fact that 
$$\bar S = S \cup \operatorname{Ac}(S),$$ 
where $\operatorname{Ac}(S)$ is the set of all accumulation points of $S$.However, from the above propositions, I get that $bar S = \operatorname{Ac}(S)$. I mean I couldn't find where should the fact that $x_n \not = x_m$ for $n \not = m$ prevent me from stating this, so my question is how can we prove that $\bar S = S \cup \operatorname{Ac}(S)$
Definition: 
$$\operatorname{Ac}(S) = \{ x\in E \mid \forall \epsilon > 0, [B(x, \epsilon) \setminus \{x\}] \cap S \not = \emptyset\}.$$
Edit:
Definition of closure;
$$\bar A = Int(A) \cup \partial (A),$$ i.e the union of the set of all interior points of $A$ and the boundary of $A$.
 A: Suppose $x\in\bar{A}$; then either $x\in\operatorname{Int}(A)$ or $x\in\partial(A)$. In any case, if $x\in A$, we have nothing to prove, so let's assume $x\notin A$ and $x\in\partial A$. By definition of boundary, each neighborhood of $x$ intersects $A$ (at a point distinct from $x$, as $x\notin A$). Therefore $x\in\operatorname{Ac}(A)$.
Similarly, suppose that $x\in A\cup\operatorname{Ac}(A)$ and that $x\notin\operatorname{Int}(A)$. Can you prove that $x\in\partial(A)$?
A: Well, take $\{x\} \subseteq \mathbb{R}$ with the euclidean metric. We have $\overline{\{x\}} = \{x\}$, but $x$ is not an accumulation point of $\{x\}$. There does exist a sequence $(x_n)_{n=1}^\infty$ such that $x_n \xrightarrow{n\to\infty} x$, namely $x_n = x$, $\forall n \in \mathbb{N}$, but you cannot obtain a sequence in $\{x\}$ with distinct elements.

The inclusion $S \cup \mathrm{Ac}\,S \subseteq \overline{S}$ is obvious.
Let's prove $\overline{S} \subseteq S \cup \mathrm{Ac}\,S$. Assume that $x \in \overline{S}$ and $x \notin S$. We shall prove that $x$ is an accumulation point of $S$.
There exists a sequence $(x_n)_{n=1}^\infty$ in $S$ which converges to $x$. Since $x \notin S$, there exists an elment $x_{p(1)}$ in the sequence such that $x_{p(1)} \ne x$.
Take $\varepsilon_1 = d(x, x_{p(1)}) > 0$. There exists $n_{\varepsilon_1} \in \mathbb{N}$, $n_{\varepsilon_1} > p(1)$ such that $n \ge n_{\varepsilon_1} \implies d(x, x_n) < \varepsilon_1$. Define $x_{p(2)} = x_{n_{\varepsilon_1}}$. It certainly holds $x_{p(2)} \ne x_{p(1)}$ since $d(x, x_{p(2)}) < d(x, x_{p(1)})$.
Take $\varepsilon_2 = d(x, x_{p(2)}) > 0$. There exists $n_{\varepsilon_2} \in \mathbb{N}$, $n_{\varepsilon_2} > p(2)$ such that $n \ge n_{\varepsilon_2} \implies d(x, x_n) < \varepsilon_2$. Define $x_{p(3)} = x_{n_{\varepsilon_2}}$. It certainly holds $x_{p(3)} \ne x_{p(1)}, x_{p(2)}$ since $d(x, x_{p(3)}) < d(x, x_{p(2)}) < d(x, x_{p(1)})$.
Now continue inductively in the same manner. You obtain a sequence $(x_{p(n)})_{n=1}^\infty$ in $S$, with all distinct elements since $d(x, x_{p(n)})$ is strictly decreasing. It also converges to $x$ as it is a subsequence of $(x_n)_{n=1}^\infty$. Hence, $x$ is an accumulation point of $S$.
