$X$ is a topological space and $x=\{x_n\}\subset X$ is a sequence. $I$ is an ideal of
y ∈ X is called an I-limit point of x if there exists a set $M = \{m_1 < m_2 < m_3 < ···\} ⊂ \mathbb N$ such that $M \notin I$ and $\lim_{k→∞} x_{m_k} = y$
y ∈ X is called an I-cluster point of x if for every open set U containing y, $\{n ∈ \mathbb N: x_n ∈ U\} \notin I$.
We denote by $I(C_x)$ and $I(L_x)$ the collections of all I-cluster points and all I-limit points of x, respectively.
Now I can easily prove that $I(L_x) ⊂ I(C_x)$.
Let $y\in I(L_x).$ Then there is a set $M\notin I$ s.t. $\lim_{k→∞} x_{m_k} = y$ i.e. $\exists k\in \mathbb N$ s,.t for any given nbd $U$ $x_{m_n}\in U \forall n\ge k\implies \{n\in\mathbb N :x_n\in U\}\notin I\implies y\in I(C_x).$
Now, whether the ideal $I$ is admissible or not has not been used in this trivial proof. Is admissibility at all required or necessary here for the this paper says "For an admissible ideal I, it is known that $I(L_x) ⊂ I(C_x)$" But this holds even without admissibility of the ideal.
An ideal $I$ of $\mathbb N$ is called admissible if every singleton $\{x\}$ is in $I.$
So, my question is why is "admissible" mentioned here?
The problem is that your proof is incorrect: if $\mathscr{I}$ is not admissible, the fact that $M\notin\mathscr{I}$ does not imply that the cofinite subsets of $M$ are not in $\mathscr{I}$, and you need that implication in order for the argument to work. Here is a concrete counterexample.
Let $\mathscr{I}=\{I\subseteq\Bbb N:0\notin I\}$; $\mathscr{I}$ is an ideal of $\Bbb N$, but it’s not admissible. Let $X=\Bbb R$ with the usual topology. Define a sequence $x=\langle x_n:n\in\Bbb N\rangle$ by
$$x_n=\begin{cases} 1,&\text{if }n=0\\ 2,&\text{otherwise}\;. \end{cases}$$
Let $m_k=k$ for $n\in\Bbb N$, and let $M=\{m_k:k\in\Bbb N\}=\Bbb N$; then $M\notin\mathscr{I}$ (since $0\in M$), and $\lim_{k\to\infty}x_{m_k}=2$, so $2\in\mathscr{I}(L_x)$. But $\{n\in\Bbb N:x_n\in(1,3)\}=\Bbb N\setminus\{0\}\in\mathscr{I}$, so $2\notin\mathscr{I}(C_x)$, and therefore $\mathscr{I}(L_x)\nsubseteq\mathscr{I}(C_x)$.