As pointed out in my comment, your argument does not work, and cannot be fixed.
Here is one way to proceed. Since $a_1$ is an adherent point of the sequence $x_n$, there is an $n_1$ such that $|x_{n_1}-a_1|<1$. Having defined $x_{n_1},\dots,x_{n_k}$, choose $n_{k+1}$ so that $n_{k+1}>n_k$ and $|x_{n_{k+1}}-a_{k+1}|<1/(k+1)$. This is possible because $a_{k+1}$ is an adherent point of the sequence $x_n$. It is now easy to check that the subsequence $x_{n_k}$ so built is Cauchy, and converges to $a$.
In slightly more detail: $x_{n_{k+1}}$ exists since there is a subsequence of $x_n$ that converges to $a_{k+1}$. We just pick a term of the sequence, with index large enough to be beyond $n_k$, and large enough so the term is close enough to $a_{k+1}$.
That the subsequence so built converges to $a$ follows from the triangle inequality: $|x_{n_k}-a|\le|x_{n_k}-a_k|+|a_k-a|$, and both terms on the right are small enough if $k$ is sufficiently large.
Yes, there are an infinite number of possible sub-sequences. However, thinking in terms of all possible sub-sequences is not going to help; there are just too many. If you want to find the limit points, first you should guess what they are by thinking about how the sequence behaves. Then, with guesses in hand, you should set about finding subsequences to verify your claims.
(1) $a_n=(-1)^n$: This sequence behaves quite simply. $a_1=-1$, $a_2=1$, $a_3=-1$, $a_4=1$, etc. In particular, since the sequence only ever takes the values $-1$ and $1$, these are the only possible limit points. Moreover, we can note that $a_{2k}=1$ and $a_{2k-1}=-1$ for all $k\in \mathbb N$; setting $m_k=2k$ and $n_k=2k-1$, we see that
$$\lim_{k\rightarrow \infty} a_{m_k}=1,\qquad \lim_{k\rightarrow \infty} a_{n_k}=-1.$$
(2) $b_n=\sin(n \pi/2)$: like with (1), we should think about the behavior of the sequence. Plugging in the first few values of $n$, we see that $b_1=1$, $b_2=0$, $b_3=-1$, $b_4=0$, $b_5=1$; in particular, these values will just cycle. This tells us that only $1,0,-1$ can possibly be limit points; moreover, since we see those numbers crop up again and again, they must be limit points. We verify this by setting $\ell_k=2k$, $m_k=4k-3$, $n_k=4k-1$. Then
$$\lim_{k\rightarrow \infty} b_{\ell_k}=0,\qquad \lim_{k\rightarrow \infty} b_{m_k}=1,\qquad \lim_{k\rightarrow \infty} b_{n_k}=-1.$$
For extra credit (and a more interesting examples):
(3) $c_n=\frac{n-1}{n}\sin(n \pi/2)$. I claim this has the same limit points as $b_n$ did. Can you prove it?
Best Answer
Clearly $a_n\in [0,1]$ and $b_n\in [-1,1]$ for each natural $n$. On the other hand, the set $\{a_n\}$ is dense in $[0,1]$ and the set $B=\{b_n\}$ is dense in $[-1,1]$ (so the sets of the limit points of $\{a_n\}$ and $\{b_n\}$ are $[0,1]$ and $[-1,1]$, respectively). Both these facts are well-known. For the first, see, for instance, my detailed recent answer here. To prove the second, note that for each $n$, $b_n=\sin 2\pi a'_n$, where $a'_n$ is $a_n$ with $\alpha$ replaced by (an other irrational number) $\alpha/2$. Since by the first fact a set $A=\{a'_n\}$ is dense in $[0,1]$ and $B$ is an image of the set $A'$ by a continuous map onto $[-1,1]$, the set $B$ is dense in $[-1,1]$.