A topological space is called a Fréchet-Urysohn space if for every $A \subset X$ and every $x \in \overline{A}$ there exists a sequence $x_1, x_2,....$ of points of $A$ converging to $x$.
So, is it right to say:
If $X$ is a Fréchet-Urysohn space , then for every cluster point $x$ of a sequence $\{x_i\}$ in $X$ there exists a subsequence of $\{x_i\}$ that converges to $x$?
Can do we use a sequential space instead of Fréchet-Urysohn space ?
I have thought more about my comment and I think I have a proof that it holds generally that in Fréchet–Urysohn space for every sequence $⟨x_n: n ∈ ω⟩$ and every its cluster point $x$ there is some subsequence $⟨x_{n_k}: k ∈ n⟩ \to x$. I'll denote sequences like $⟨x_n: x ∈ ω⟩$ and their ranges like $\{x_n: n ∈ ω\}$.
Let's take $I := \{n ∈ ω: ∀U \text{ open } x ∈ U \implies x_n ∈ U\}$ i.e. indices of sequence elements such that $x$ cannot be separated from them. If $I$ is cofinal in $ω$ then we can take a subsequence converging to $x$. If not then it's finite so we can strip an initial segment of the sequence so we can assume that $x$ can be separated from any $x_n$.
$x$ is cluster point, specially $x ∈ \overline{\{x_n: n ∈ ω\}}$. Because $X$ is Fréchet–Urysohn, there is some $⟨y_m: m ∈ ω⟩ \to x$ such that $\{y_m: m ∈ ω\} ⊆ \{x_n: n ∈ ω\}$. Let's take $n_0$ such that $x_{n_0} = y_0$ and $m_0 = 0$. Now if we have $n_j, m_j$, $j ≤ k$, we will take $n_{k + 1} > n_k$ and $m_{k + 1} > m_k$ such that $x_{n_{k + 1}} = y_{m_{k + 1}}$. We can do this or else $(∀n > n_k): x_n ∈ \{y_m: m ≤ m_k\}$ and so $x$ is in closure of finite number of elements of original sequence and hence cannot be separated from all of them which is contradiction with our assumption. So by this inductive construction we obtain $⟨x_{n_k}: k ∈ ω⟩ = ⟨y_{m_k}: k ∈ ω⟩$ a subsequence of $⟨x_n: n ∈ ω⟩$ which is also a subsequence of $⟨y_m: m ∈ ω⟩$ and so converges to $x$.
It is not enough to have a sequential space. For example, the Arens space is a sequential space but not a Fréchet-Urysohn space. To define it, take a topological sum of convergent sequences $(x_{n,k})_k\to x_n$ for all $n\in\Bbb N$ plus a convergent sequence $(x_{*,k})_k\to x_*$. All points $x_{n,k},\ x_{*,k}$ are isolated. Then take the quotient by identifying $x_{*,n}\sim x_n$, so that the limits $(x_n)$ themselves form a sequence converging to $x_*.$ This space, call it $A$, is called the Arens space. By construction a neighborhood of $x_*$ contains almost all points from almost all sequences. The Arens space is sequential since it's the quotient of a first-countable space.
We will now construct a sequence with a cluster point which is not the limit of a subsequence. The subspace $B=A-\{x_n\mid n\in\Bbb N\}$ is called the Arens-Fort space. We can construct a sequence $(b_n)_n$ which exhausts $B-\{x_*\}$ going through the points in a zig-zag way: $$x_{1,1}\to x_{1,2}\to x_{2,1}\to x_{3,1}\to x_{2,2}\to x_{1,3}\to x_{1,4}\to x_{2,3}\to x_{3,2}\to x_{4,1}\to...$$ Every neighborhood of $x_*$ has infinitely many elements from $B$, so $x_*$ is a cluster point of $(b_n).$ On the other hand, there is no subsequence of $(b_n)$ converging to $x_*.$ In fact, one can show that no sequence in $B-\{x_*\}$ converges to $x_*.$ But $x_*$ is in the closure of $B-\{x_*\}$, so this means that the Arens space is not a Fréchet-Urysohn space.
Let's see what happens in Fréchet-Urysohn spaces. Let $c$ be a cluster point of the sequence $(x_n)_n$ and $S$ denote the range of $(x_n).$
If some neighborhood of $c$ contained only finitely many elements from $S$, then some of them, call them $z_1,...,z_l$, are assumed infinitely often by the sequence since $c$ is a cluster point. If for each $i=1,...,l$ there were some neighborhood $U_i$ of $c$ with $z_i\notin U_i$, then the intersection of all $U_i$ would be a neighborhood containing only points of $S$ that are met finitely often by $(x_n)$, contradicting $c$ is cluster point. On the other hand, if one $z_j$ is in each neighborhood of $c$, then the constant subsequence at that point converges to $c$ and we are finished.
Now we consider the case that each $x\in S$ which is assumed by the sequence infinitely often is excluded from some neighborhood of $c.$ If infinitely many elements of $S$ were contained in each neighborhood of $c$, then they would form a convergent subsequence. So only a finite number of elements in $S$ are contained in the intersection of all neighbourhoods of $c$. But they are are all met only finitely often, so there is a tail $\mathcal T=(x_n)_{n>n_0}$, which still has $c$ as cluster point, but every $x_n$ in $\cal T$ can be excluded from some neighborhood of $c$. As $c$ is in the closure of $T=\{x_n\mid n>n_0\}$ and $X$ is Fréchet-Urysohn, there is a sequence $(y_n)_n$ in $T$ converging to $c$. Since each point in $T$ can be separated from $c$, $(y_n)$ must have infinitely many distinct values, so we can extract a subsequence of $(y_n)$ which is also a subsequence of $(x_n)_n$. (For the details of this construction see user87690's answer.)
So a Fréchet-Urysohn space has the property that each cluster point of a sequence is also the limit of a convergent subsequence.
