Fréchet-Urysohn space We know that if $p$ is the limit of a subsequence $(x_{n_{k}})$ of the sequence $(x_n)$ in $X$, then $p$ is a cluster point of the sequence.
For a sequential space, it does not hold that a cluster point of a sequence is the limit of a subsequence of it.
If we have a sequence $(x_n)$ in a  Fréchet-Urysohn space with a cluster point p, will this give us a convergent subsequence to p? How?
 A: Yes, it will. By definition a Fréchet-Urysohn space is one in which the sequential closure of a set is equal to the closure of the set. Suppose that $X$ is Fréchet-Urysohn, $\sigma=\langle x_n:n\in\omega\rangle$ is a sequence in $X$, and $p$ is a cluster point of $\sigma$. 
Let $\mathscr{B}_p$ be the set of open nbhds of $p$, and let $C_p=\bigcap\mathscr{B}_p$; $C_p=\{p\}$ if $X$ is $T_1$, but if $X$ is not $T_1$, it’s possible that $C_p\supsetneqq\{p\}$. Let $M=\{n\in\omega:x_n\in C_p\}$. If $M$ is infinite, $\langle x_n:n\in M\rangle$ is a subsequence of $\sigma$ converging to $p$, so we may assume that $M$ is finite. Let $m=1+\max M$, and let $\sigma'=\langle x_n:n\ge m\rangle$. Let $U$ be any open nbhd of $p$, and let $\ell\in\omega$; $p$ is a cluster point of $\sigma$, so there is an $n\ge\max\{m,\ell\}$ such that $x_n\in U$, so $p$ is a cluster point of $\sigma'$, and for all $n\ge m$ we have $x_n\notin C_p$. Thus, we may as well replace $\sigma$ by $\sigma'$ if necessary and assume that $x_n\notin C_p$ for all $n\in\omega$.
Let $A=\{x_n:n\in\omega\}$. $A\cap C_p=\varnothing$, so $p\notin A$. On the other hand, the fact that $p$ is a cluster point of the sequence $\sigma$ implies that $p$ is a limit point (or accumulation point) of the set $A$ and hence that $p\in\operatorname{cl}A$. $X$ is Fréchet-Urysohn, so $\operatorname{cl}A$ is the sequential closure of $A$; this means that there is a sequence $\langle a_k:k\in\omega\rangle$ in $A$ that converges to $p$. For each $k\in\omega$ there is an $n_k\in\omega$ such that $a_k=x_{n_k}$, and the sequence $\langle x_{n_k}:k\in\omega\rangle$ converges to $p$. 
This is almost what we want, but not quite: $\langle x_{n_k}:k\in\omega\rangle$ might not be a subsequence of $\sigma$, because the sequence $\langle n_k:k\in\omega\rangle$ may not be increasing. I’ll leave it to you to show that $\langle n_k:k\in\omega\rangle$ has a strictly increasing subsequence; you’ll need to use the fact that for each $k\in\omega$ there is an open nbhd of $p$ that does not contain $x_{n_k}$.
Note: If we assume that $X$ is $T_1$, the argument can be simplified a bit; making $\sigma$ avoid the set $C_p$ is to take care of the possibility that $X$ is not $T_1$. It’s necessary to look at the set $A$ because the definition of Fréchet-Urysohn spaces imposes a condition on the closures of sets: I use the fact that $p$ is in the closure of the set $A$ to conclude that $p$ is in the sequential closure of $A$ and hence that some sequence in $A$ converges to $p$. To complete the proof one must then show that this sequence has a subsequence that is also a subsequence of the original $\sigma$; that’s the bit that I left for you.
