Converging sequences in a given topological space. let $A=[0,1]$, and $\tau=[0,1]\mathbin{\vcenter{\hbox{$\scriptscriptstyle\setminus$}}}{a_n}$ where $a_n$ is a finite or infinite sequence in A. Now let $T=(A,\tau)$, the question is what type of sequences converge $x_n$ in this topological space? Apparently the answer is $x_n$ must be constant after some $n=N$ (told by the lecturer), and I just don't quite see this, I think $x_n$ just need to not contain $a_i$ after some $N=n$. But I don't know.
This was used to demonstrate that not all points of closure in a topological space have sequences converging to them: so following his argument, let $B=(0,1], \overline{B}=A$ but no sequences converge to $0$ since $0\notin B$. 
 A: I suspect that you’ve misunderstood the definition of $\tau$, and that $\tau$ is supposed to be defined as follows:
$$\tau=\{\varnothing\}\cup\{[0,1]\setminus C:C\text{ is finite or countably infinite}\}\;.$$
In other words, the open sets are $\varnothing$, the complements of finite sets, and the complements of countably infinite sets. With this topology, $[0,1]$ has the desired property.
Suppose that $\langle x_n:n\in\Bbb N\rangle$ is a sequence in $[0,1]$ converging to some $x\in[0,1]$ in this topology. Let $M=\{n\in\Bbb N:x_n\ne x\}$, and let $C=\{x_n:n\in M\}$. Finally, let $U=[0,1]\setminus C$. Then $[0,1]\setminus U=C$, and $C$ is finite or countably infinite, so $U\in\tau$. Moreover, $x\notin C$, so $x\in U$; in other words, $U$ is an open nbhd of $x$. And $\langle x_n:n\in\Bbb N\rangle$ converges to $x$, so there must be an $m\in\Bbb N$ such that $x_n\in U$ whenever $n\ge m$. That means that if $n\ge m$, then $x_n\notin C$, so $n\notin M$, and therefore, by the definition of $M$, $x_n=x$. In other words, $x_n=x$ for all $n\ge m$, and the sequence is indeed eventually constant at $x$.
