Show that a sequence in $X$ converges $⇔$ it has a single cluster point The problem says
Let $X$ be a countably compact $T 2$ -space satisfying the first axiom of
countability. Show that a sequence in $X$ converges $⇔$ it has a single
cluster point.
If x is the accumulation point. Can I make my succession of smaller and smaller neighborhoods of x?
 A: Let $(x_n)_n$ be a sequence in $X$.
If $x_n\to x$, then $x$ is a cluster point of $(x_n)$. To see that this is the only one, suppose $y$ is another cluster point of $(x_n)$ and fix a countable base $\{U_n \mid n\in\mathbb{N}\}$ of neighborhoods of $y$ such that $U_{n+1}\subseteq U_n$ for each $n$. Choose $n_1$ such that $x_{n_1}\in U_1$, and having found $n_1<\cdots<n_k$, choose $n_{k+1}>n_k$ such that $x_{n_{k+1}}\in U_{k+1}$. This defines a subsequence $(x_{n_k})_k$ of $(x_n)$ such that $x_{n_k}\to y$. Then we also have $x_{n_k}\to x$, so the fact that $X$ is Hausdorff implies that $y=x$. Therefore $(x_n)$ has only one cluster point.
Conversely suppose that $x$ is the only cluster point of $(x_n)_n$ and assume by contradiction that $x_n\not\to x$. Then there is neighborhood $U$ of $x$ such that $\mathbb{N}_1:=\{n\in\mathbb{N} \mid x_n\not\in U\}$ is an infinite set. Consider the subsequence $(x_n)_{n\in\mathbb{N}_1}$ of $(x_n)_{n\in\mathbb{N}}$. Since $X$ is countably compact, this subsequence must have a cluster point $y$. Hence $y$ is a cluster point of the original sequence. As $x$ is not a cluster point of $(x_n)_{n\in\mathbb{N}_1}$, we have $x\ne y$. This contradicts the assumption that there is only one cluster point. We conclude $x_n\to x$.
