Sequence is union of convergent subsequences. Are the limits of the subsequences the only cluster points the sequence have? Given a sequence $(a_n)_n$, when one is asked to find all the cluster points of this sequence in $(\mathbb{R},d_E)$, one can find convergent subsequences and their limit is a cluster point. However, to make sure we have found all the cluster points, does it suffice to show that if we can write the (range of the) sequence as finite union of (ranges of) convergent subsequences, then the limits of the subsequences are ALL cluster points.
I managed to prove that all these limits of subsequences are cluster points, but I don't know how to prove that these are all possible cluster points.
Formally, my question is: Is the following statement true, and why:

Suppose $(a_n)_n$ is a sequence and $$\{a_n|n \in \mathbb{N}\} =
\bigcup_{i=1}^m \{a_{{ki}_n}|n \in \mathbb{N}\}$$ with $a_{{ki}_n} \to
 a_i$. Then $\{a_1, \dots , a_m\}$ is the set of all cluster points of
  $(a_n)_n$

My definition of cluster point:

$x$ cluster point of a sequence $(a_n)_n$ in a metric space $(X,d)
 \iff \forall \epsilon > 0: \forall n \in \mathbb{N}: \exists m > n:
 d(x_m,x)  <\epsilon$

Can someone hint me into the right direction? (I don't even know if this statement is true or false actually)
 A: We need to cover the set of indices, not the set of terms of the sequence. As written it is not ruled out that some point $x$ is attained infinitely often by the sequence $(a_n)$, but each of the subsequence attains $x$ only a finite number of times. Then $x$ is a cluster point of $(a_n)$, but it need not be the limit of any of the subsequences.
If the indices used in the subsequences cover $\mathbb{N}$ except perhaps for finitely many $n$, then the conclusion holds.
Let's use a different notation for subsequences which I find clearer. Given a sequence $(a_n)_{n\in\mathbb{N}}$, its subsequences are in bijective correspondence with the strictly increasing maps $\mathbb{N}\to \mathbb{N}$. Thus we can write a subsequence of $(a_n)_{n\in\mathbb{N}}$ as $(a_{\sigma(n)})_{n\in\mathbb{N}}$.
So, if we have a real sequence(1) $(a_n)_{n\in\mathbb{N}}$ and $m$ convergent subsequences $\bigl(a_{\sigma_i(n)}\bigr)$ such that
$$E := \mathbb{N}\setminus \bigcup_{i = 1}^m \sigma_i(\mathbb{N})$$
is finite. Then
$$L := \Bigl\{ \lim_{n\to\infty} a_{\sigma_i(n)} : 1 \leqslant i \leqslant m\Bigr\}$$
is the set of all cluster points of $(a_n)_{n\in \mathbb{N}}$.
Proof: I omit the proof that the limit of a subsequence is a cluster point of the sequence and only show that $c \notin L$ implies that $c$ is not a cluster point of $(a_n)$. Fix $c\notin L$. Let
$$\delta = \min \Bigl\{ \bigl\lvert c - \lim_{n\to\infty} a_{\sigma_i(n)}\bigr\rvert : 1 \leqslant i \leqslant m\Bigr\}.$$
As the minimum of a finite set of strictly positive numbers, $\delta$ is itself strictly positive. Let $\varepsilon = \delta/2$. By the assumed convergence of the subsequences, for each $i$ there is an $n_i$ such that
$$n \geqslant n_i \implies \bigl\lvert a_{\sigma_i(n)} - \lim_{k\to\infty} a_{\sigma_i(k)}\bigr\rvert < \varepsilon.$$
Define
$$N = \max \Bigl(\bigl\{ \sigma_i(n_i) : 1 \leqslant i \leqslant m\bigr\} \cup \{n_E\}\Bigr)$$
where $n_E = 0$ if $E = \varnothing$ and $n_E = 1 + \max E$ if $E\neq \varnothing$.
Then we have $\lvert a_n - c\rvert > \varepsilon$ for $n \geqslant N$. For, if $n \geqslant N$, then $n \notin E$, so there is an $i$ with $n \in \sigma_i(\mathbb{N})$. Since $\sigma_i(n_i) \leqslant N \leqslant n$, it follows that $k = \sigma_i^{-1}(n) \geqslant n_i$ by the monotonicity of $\sigma_i$. Hence, by the choice of $n_i$, we have
$$\bigl\lvert a_n - \lim_{k\to\infty} a_{\sigma_i(k)}\bigr\rvert < \varepsilon$$
and consequently
$$\lvert a_n - c\rvert \geqslant \bigl\lvert c - \lim_{k\to\infty} a_{\sigma_i(k)}\bigr\rvert - \bigl\lvert a_n - \lim_{k\to\infty} a_{\sigma_i(k)}\bigr\rvert > 2\varepsilon - \varepsilon.$$
Thus $c$ is not a cluster point of $(a_n)$.

(1) It's not important that it's a sequence in $\mathbb{R}$, any metric space - in fact any first countable Hausdorff space - would work, with [almost] the same proof.
