Does "topological invariance" of the set of limit points or accumulation points imply the "eventual constantness" of a sequence? Problem

Let $(X,\tau_0)$ be a topological space with $X(\ne \emptyset)$. Let $(x_n)_{n\in\mathbb{N}}\in X$ and let the set of all adherent points of $(x_n)_{n\in\mathbb{N}}\in X$ with respect to the toplogy $\tau_0$ be $\mathscr{L}(\tau_0)$. If for any topology (not the indiscreet and discreet ones) $\tau$ on $X$ we have $\mathscr{L}(\tau)=\mathscr{L}(\tau_0)$ then prove or disprove that the sequence $(x_n)_{n\in\mathbb{N}}$ is ultimately constant. Will your conclusion be same if $\mathscr{L}(\tau)$ would denote the set of all accumulation points? 

I have tried to think over the problems for quite sometime but without any progress so far. The almost trivial result that I could prove was if the set $\mathscr{L}(\tau)(\ne \emptyset)$ denotes the set of all limits of $(x_n)_{n\in\mathbb{N}}$ then the conclusion of the problem holds, i.e., the sequence is ultimately constant. 
Can anyone help me in solving the two problems?

Definition of Accumulation Point which I am using: A point $x\in X$ is said to be an accumulation point of the sequence $(x_n)_{n\in\mathbb{N}}\in X$ if there existis a subsequence of $(x_n)_{n\in\mathbb{N}}$, denoted by $(x_{n_k})_{k\in\mathbb{N}}$ such that $x_{n_k}\ne x$ for all $k$ and it converges to $x$.
Definition of Adherent Point which I am using: A point $x\in X$ is said to be an adherent point of the sequence $(x_n)_{n\in\mathbb{N}}\in X$ if there existis a subsequence of $(x_n)_{n\in\mathbb{N}}$, denoted by $(x_{n_k})_{k\in\mathbb{N}}$ such that it converges to $x$.
 A: Counterexample: Consider the set $X=\mathbb{N}$ and two 'near'-indiscrete topologies, $\tau_i$ and $\tau_i^*$. We will arrange that each of them has two nonempty proper subsets, the larger such subsets of $X$, $U_i$ for $\tau_i$, and $U_i^*$ for $\tau_i^*$, chosen so that the given subsets contains all but one natural number; $n_i$ for $\tau_i$ and $n_i^*$ for $\tau_i^*$, furthermore, we will do so, in a way that $n_i\neq n_i^*$, whereas the smaller sets will simply be the complements of of $U_i$ and $U_i^*$ If the sequence $(x_n)_{n\in\mathbb{N}}$ does not contain infinitely many copies of either of $n_i$ or $n_i^*$ (note: this means that every subsequence has limit(s), inside both of $U_i$ and $U_i^*$), then we have found topologies such that $\mathscr{L}(\tau)\neq\mathscr{L}(\tau_0)$, for the set of adherent points for $\tau_i$ and $\tau_i^*$ will be $U_i$ and $U_i^*$ respectively. Hence in order to have $\mathscr{L}(\tau_i)=\mathscr{L}(\tau_i^*)$, we need that the sequence contains infinitely many copies of every element, else we can successfully construct the previously described $U_i$ and $U_i^*$ which lead to $\mathscr{L}(\tau)\neq\mathscr{L}(\tau_0)$. Any sequence $(x_n)_{n\in\mathbb{N}}$ which contains infinitely many copies of more than one point cannot be eventually constant because that implies that, after some finite number of terms, all remaining terms are the same point, which is a contradiction (implies every other point must occur only finitely many times). Hence, for $X=\mathbb{N}$, $(x_n)_{n\in\mathbb{N}}$ cannot be ultimately constant.
If we require that the points be accumulation points rather than merely adherent points, now we consider $X=\mathbb{R}$ and define 'near'-indiscrete topologies similarly as before. Since, $\mathbb{R}$ is uncountable, we are guaranteed that we can find, $r_i\neq r_i^*$ (in $\mathbb{R}$) not in $(x_n)_{n\in\mathbb{N}}$ to use to construct $U_i$ and $U_i^*$ as before. Note that $r_i\in U_i^*$ and $r_i^*\in U_i$. Now in $\tau_i$, since, $\{r_i\}$ is open, we know that $r_i$ is not an accumulation point of $(x_n)_{n\in\mathbb{N}}$ for the topology $\tau_i$. At the same time, we have that $r_i^*$ is an accumulation point because it is not in $(x_n)_{n\in\mathbb{N}}$ but $(x_n)_{n\in\mathbb{N}}$ is eventually entirely within every neighborhood of $r_i^*$ (smallest of which is $U_i$). By symmetric arguments, we have that, for the topology $\tau_i^*$, $r_i^*$ is not an accumulation point of $(x_n)_{n\in\mathbb{N}}$ because $\{r_i^*\}$ is open, but also, that $r_i$ is an accumulation point of $(x_n)_{n\in\mathbb{N}}$ because $(x_n)_{n\in\mathbb{N}}$ is eventually within every neighborhood of $r_i$ (smallest being $U_i^*$). Therefore, we have found topologies for $X=\mathbb{R}$ such that $\mathscr{L}(\tau)\neq\mathscr{L}(\tau_0)$. (Hence the conclusion does not change!)
