Let $(x_n)_{n\in \mathbb N}$ be a sequence of reals in $[0,1]$ indexed on the positive integers.
Let $L\subset [0,1]$ be the limit points of $(x_n)$, in the sense that $l\in L$ if and only if there exists an infinite subsequence of $(x_n)$ converging to $l$.
Can we always construct a partition $(N_l)_{l\in L}$, indexed in the limit points $L$, such that for any $l\in L$, the subsequence $$(x_n)_{n\in N_l} \to l.$$
No. You can have a sequence with uncountably many limit points, in fact every point of $[0,1]$ could be a limit point. For example, this is the case with the sequence $0/1, 1/1, 0/2, 1/2, 2/2, 0/3, 1/3, 2/3, 3/3, \ldots$. You can't have uncountably many disjoint nonempty subsets of a countable set.
[EDIT] If you choose a set $S = \{s_1, s_2, \ldots\}$ of limit points that is countably infinite, then yes: it is possible to partition $\mathbb N$ into disjoint subsequences $\mathbb N_i$, with $x_n$ for $n \in N_i$ converging to $s_i$.
You can construct the partition inductively. Suppose we have $\mathbb N_1, \ldots, \mathbb N_k$, the corresponding sequences $\langle x_n : n \in N_i\rangle$ converging to $s_i$. There is $\epsilon > 0$ such that
$|s_{k+1} - s_i| > 2 \epsilon$ for $i \le k$. For each $i \le k$, there are at most finitely many $x_n$ with $n \in \mathbb N_i$ and $|x_n - s_{k+1}| < \epsilon$. Thus there is $\delta > 0$ such that for every $n \in \mathbb N_i, i \le k$, either $x_n = s_{k+1}$ or $|x_n - s_{k+1}| > \delta$, and only finitely many such $n$ with $x_n = s_{k+1}$: take $m$ be greater than the maximum of that finite set of $n$. Let $\mathbb N_{k+1}$ contain $k+1$ if that is not already in one of the $\mathbb N_{i}$, and then an increasing sequence $n_1 < n_2 < \ldots$ such that $n_1 > m$ and $|x_{n_i} - s_{k+1}| < 2^{-i} \delta$. Since $s_{k+1}$ is a limit point of your sequence, it is possible to choose this. By construction it is disjoint from $\mathbb N_i$ for $i \le k$, and every member of $\mathbb N$ is in some $\mathbb N_i$.