Let $(a_n)$ be a sequence and consider subsequences $(a_{{p_k}_n})$ $k=1,2,3,......$, that means we consider infinitely many subsequences of $a_n$. Suppose these subsequences are pairwise disjoint and form the sequences $(n)$. If $S, S_{p_1}, S_{p_2}, ... $ are set of limit points of $(a_n), (a_{{p_k}_n}) $, we want to prove that
$$ S = \bigcup_{k=1}^{\infty} S_{p_k} $$
And use this to prove that if every subsequences $(a_{{p_k}_n}) $ converges to $L$, then $a_n \to L$.
Attempt:
If $x \in \bigcup S_{p_k}$, then $x \in S_{p_k}$ for some $k$ so there exists a subsequence of the subsequence $(a_{p_{k_n}})$ of $(a_n)$ that converges to $x$ by definition of limit point. But, this subsequence of the subsequence is itself a subsequence of $(a_n)$. In other words, we have found a subsequence of $(a_n)$ that converges to $x$: $x \in S$. We have $ \boxed{ \bigcup_{k \geq 1} S_{p_k} \subset S }$
It is the other direction that it hard: We can instead prove that $( \bigcup S_{p_k} )^c \subset S^c \iff \bigcap S^c_{p_k} \subset S^c$. So if $x \in \bigcap S^c_{p_k} $ then $x$ is in every $S_{p_k}^c $.
Now, $x \notin S_{p_k}$ means that $x$ is not a limit point of $S_{p_k}$. How can we negate the statement of not being a limit point: Is it that there is no subsequnece that converges to $x$? or is it that every subsequence not converges to $x$ ?
The statement to be proved is false. Let $\mathbb{P}$ the set of primes and WLOG suppose that $L=0$, then for every $p\in \mathbb{P}$ set a subsequence defined by $a_{p^k}:=1/k$ for $k\in \mathbb{N}\setminus\{0\}$, and let any other $a_m:=0$ if $m\neq p^k$ for some prime $p$ and some $k\in \mathbb{N}\setminus\{0\}$.
Then by construction the subsequences $(a_{p^k})_{k\in \mathbb{N}\setminus\{0\}}$ are all mutually disjoint for distinct $p\in \mathbb{P}$ and converges to zero, however each subsequence $(a_{p^k})_{p\in \mathbb{P}}$ is constantly $1/k$, so $1/k$ is a limit point of $(a_n)$ for each $k\in \mathbb{N}\setminus\{0\}$.