[proofplan]
We translate convergence in a [topological space](/page/Topological%20Space) into eventual membership in every open neighbourhood of the candidate limit $x$. For the forward implication, the cofinite [open set](/page/Open%20Set) $X\setminus\{y\}$ excludes a fixed point $y\neq x$, so eventual membership in this neighbourhood forces the sequence to hit $y$ only finitely often. For the reverse implication, an arbitrary neighbourhood $U$ of $x$ has finite complement $F=X\setminus U\subset X\setminus\{x\}$; the assumed finiteness for each point of $F$ implies that the sequence eventually avoids all of $F$, hence is eventually in $U$.
[/proofplan]
[step:Use convergence to show that every point different from $x$ is hit only finitely often]
Assume that $(x_n)_{n\in\mathbb{N}}$ converges to $x$ in the [cofinite topology](/page/Cofinite%20Topology) on $X$. Fix $y\in X\setminus\{x\}$. Define
\begin{align*}
U_y:=X\setminus\{y\}.
\end{align*}
Since $X\setminus U_y=\{y\}$ is finite, $U_y$ is open in the cofinite topology. Also $x\in U_y$ because $y\neq x$, so $U_y$ is an open neighbourhood of $x$.
By convergence, there exists $N\in\mathbb{N}$ such that $x_n\in U_y$ for every $n\in\mathbb{N}$ with $n\ge N$. Therefore $x_n\neq y$ for every $n\ge N$, and hence
\begin{align*}
\{n\in\mathbb{N}:x_n=y\}\subset \{1,\dots,N-1\}.
\end{align*}
The set on the right is finite, so $\{n\in\mathbb{N}:x_n=y\}$ is finite.
[guided]
Assume that $(x_n)_{n\in\mathbb{N}}$ converges to $x$ in the cofinite topology on $X$. To prove the required finiteness condition, we must fix an arbitrary point $y\in X\setminus\{x\}$ and show that the sequence can equal $y$ only finitely many times.
Define
\begin{align*}
U_y:=X\setminus\{y\}.
\end{align*}
This is the right neighbourhood to test because it contains the proposed limit $x$ but excludes exactly the point $y$. Since $X\setminus U_y=\{y\}$ is finite, the definition of the cofinite topology says that $U_y$ is open. Since $y\neq x$, we also have $x\in U_y$, so $U_y$ is an open neighbourhood of $x$.
Now apply the definition of convergence of a sequence in a topological space: for every open neighbourhood of the limit, all sufficiently late terms of the sequence must lie in that neighbourhood. Applying this to the open neighbourhood $U_y$, there exists $N\in\mathbb{N}$ such that $x_n\in U_y$ for every $n\in\mathbb{N}$ with $n\ge N$. Since membership in $U_y=X\setminus\{y\}$ is exactly the condition of being different from $y$, this says that $x_n\neq y$ for every $n\ge N$.
Thus any index $n$ for which $x_n=y$ must satisfy $n<N$. Hence
\begin{align*}
\{n\in\mathbb{N}:x_n=y\}\subset \{1,\dots,N-1\}.
\end{align*}
The set $\{1,\dots,N-1\}$ is finite, and every subset of a finite set is finite. Therefore $\{n\in\mathbb{N}:x_n=y\}$ is finite.
[/guided]
[/step]
[step:Use the pointwise finiteness condition to get eventual membership in every neighbourhood of $x$]
Assume conversely that for every $y\in X\setminus\{x\}$, the set
\begin{align*}
A_y:=\{n\in\mathbb{N}:x_n=y\}
\end{align*}
is finite. Let $U\subset X$ be an open neighbourhood of $x$ in the cofinite topology. Define the complement
\begin{align*}
F:=X\setminus U.
\end{align*}
Since $U$ is open in the cofinite topology and $U\neq\varnothing$ because $x\in U$, the set $F$ is finite. Also $x\notin F$, because $x\in U$, so $F\subset X\setminus\{x\}$.
For each $y\in F$, the set $A_y$ is finite by hypothesis. Define
\begin{align*}
A_F:=\bigcup_{y\in F}A_y.
\end{align*}
If $F=\varnothing$, then $A_F=\varnothing$, so $A_F$ is finite. If $F\neq\varnothing$, then $A_F$ is a finite union of finite sets, hence finite. In either case, $A_F$ is finite.
Since $A_F$ is a finite subset of $\mathbb{N}$, there exists $N\in\mathbb{N}$ such that every $n\in A_F$ satisfies $n<N$. For every $n\ge N$, we then have $n\notin A_F$, so $x_n\notin F$. Therefore $x_n\in X\setminus F=U$ for every $n\ge N$.
[/step]
[step:Conclude convergence to $x$ from eventual membership in arbitrary neighbourhoods]
The preceding step proves that for every open neighbourhood $U$ of $x$, there exists $N\in\mathbb{N}$ such that $x_n\in U$ for every $n\ge N$. This is exactly the definition of convergence of $(x_n)_{n\in\mathbb{N}}$ to $x$ in the cofinite topology on $X$. Combining this with the first step proves the equivalence.
[/step]
