Limit point for the direct limit of tower of topological spaces. 
We consider a direct limit of a tower $X_1\subset\cdots\subset X_n$ of spaces, where each $X_n$ is a subspace of $X_{n+1}$. The direct limit is $X_{\infty}:=\cup_n X_n$ endowed with the topology $\mathcal{T}_{\infty}$ defined as follows:
$U\subset X_{\infty}$ is open if and only if $\forall n\in \Bbb{N},\quad U\cap X_n$ is open on $\mathcal{T_n}.$
Now I have a strictly increasing function $\iota$ from $\Bbb{N}$ to $\Bbb{N}$ and for all $n\in \Bbb{N}$ let $x_n$ be an element of $X_{\iota(n)}$ and if $\iota(n)>0$ assume that $x_n\notin X_{\iota(n)-1}.$
I would like to show that the sequence $x_n$ cannot have a limit point.

By limit point I mean the following definition:
$l$ is a limit point of $x_n$ if $$(\forall V\in\mathcal{B}(l))\quad(\forall n_0)\quad(\exists n\ge n_0); x_n\in V$$
I tried by arguing by contradicion:
If $x_n$ admits a limit point $l$ then I can find an integer $m$ such that for all $m_0\in\Bbb{N}:\quad m\ge m_0\implies x_m\in V$ where $V$ is any neighborhood of $l.$
But a neighborhood of $l\in X_{\infty}$, by definition, is a set $V\subset X_{\infty}$ such that there exist $U\in \mathcal{T}_{\infty}; l\in U,\quad U\subset V.$
I was thinking that I may take an element $m_0$ such that it cannot exist a open set $U$ because it will be violating the condition "$\iota(n)>0$ assume that $x_n\notin X_{\iota(n)-1}.$".
I didn't succeed, I am a bit lost with all the conditions.
 A: To simplify things, we consider a sequence $(x_n)_{n \in \mathbb N}$ in $X_\infty$ such that $x_n \in X_n \setminus X_{n-1}$. We also assume that each $X_n$ is a $T_1$-space. We want to show that the sequence has no limit points.
Define the set $Y = \{ x_1, x_2, \dots \}$ and consider $z \in X_\infty$. By discarding finitely many elements of the sequence if necessary we can assume that $z \notin Y$ and that $z \in X_1$. We will construct iteratively an open neighborhood $U$ of $z$ in $X_\infty$ that does not intersect $Y$. Choose $U_1 \subseteq X_1$ open such that $z \in U$ but $x_1 \notin U_1$, which is possible because $X_1$ is a $T_1$-space. Because $X_1 \subseteq X_{2}$ carries the subspace topology, 
$$U_1 = V_{2} \cap X_1\,,$$ 
for some $V_{2} \subseteq X_{2}$ open. Define $U_{2} = V_{2} \setminus \{x_{2}\}$ which is also open. Then also $U_1 = U_{2} \cap X_1$ and we proceed in the same way to construct $U_3$, $U_4$, etc. Now set
$$ U = \bigcup_n U_n \subseteq X_\infty\,. $$
Then $x_n \notin U$ and $U \cap X_n = U_n$ is open. Thus $U$ is open in $X_\infty$ as required.
Note: The assumption that each $X_n$ is a $T_1$-space is necessary. To see this consider $X_n = \{1,\dots,n\}$ with the topology $\mathcal T_n = \{ \emptyset, X_n \}$ and the sequence $x_n = n$. Then $X_\infty = \mathbb N$ and $\mathcal T_\infty = \{ \emptyset, \mathbb N \}$. In particular, every element of $\mathbb N$ is a limit point of the sequence.
Edit: Replaced wrong counterexample with (hopefully) correct proof.
