Let $x_n=n, \space \space n\geq 1 $ be a sequence in the topological space $X=\Bbb N$ and $T$ be the collection of all subsets whose compliment is finite and $\emptyset $.
Now i believe that the sequence converges to every point in X because for each set there is a largest element not in that set. for example i can write down an infinite number of sets containing 1 this seems to imply that the sequence approach one. how do i prove this?
Edit:
For every open set A containing $1$ there is a greatest element not in A call it d let $N= d+1$ then for all $n \geq N$ the sequence $x_n$ is entirely contained in A so the sequence converges to 1.
However 1 was an arbitrarily chosen element the same statement works for any number you wish to place there from $\Bbb N$ so i believe $x_n$ converges to every point.
You are correct for your chosen sequence $x_n=n$, and can generalize what you've found somewhat.
Let $X$ be any infinite set, $\mathcal{T}_{cf}$ be the cofinite topology as you've defined it, $\{x_n\}$ be any sequence that assumes infinitely many values, and $x^*$ be an arbitrary point in $X$. Then for every $A \in \mathcal{T}_{cf}$ with $x^* \in A$, we have that $X-A$ contains finitely many elements. Since $\{x_n\}$ assumes infintely many values, we can avoid all of the finitely many elements in $X-A$ by picking a significantly large $N \in \mathbb{N}$ so that for $n \geq N$ we have $x_n \not\in X-A$. That is, for $n \geq N$ we have $x_n \in A$. Since infinitely many elements of the sequence are in $A$, we have $x_n \to x^*$.
See the answer by José Carlos Santos for what happens if your sequence assumes finitely many values.