general-topology
Here are the definitions:
Fréchet-Urysohn space: A topological space $ X $ where for every $ A \subseteq X $ and every $ x \in \text{cl}(A) $, there exists a sequence $ (x_{n})_{n \in \mathbb{N}} $ in $ A $ converging to $ x $.
Sequential space: A topological space $ X $ where a set $ A \subseteq X $ is closed iff $ A $ contains the limit points of every sequence contained in it.
As the title explains, I would like to know the difference between them. Thanks for any help.
I will make use here of the notation introduced in my previous post Topology of the space $\mathcal{D}(\Omega)$ of test functions. We shall show that $\mathcal{D}(\Omega)$ is not a sequential space, so that the answer to both questions is negative, as announced.
Take $V$ as in my answer to the post Topology of the space $\mathcal{D}(\Omega)$ of test functions and let $A$ be the complement of $V$. Then the argument given in that answer shows that $0$ is a limit point of $A$, so $A$ is not closed. Anyhow, $A$ is sequentially closed, as we shall now show.
Suppose that $f \in V$ and that $(f_j)$ is a sequence in $\mathcal{D}(\Omega)$ converging to $f$. Then by the characterization of converging sequences in $\mathcal{D}(\Omega)$ (see e.g. Theorem (6.5) in Rudin, Functional Analysis, 2nd Edition), we know that:
(i) there is a compact set $K$ contained in $\Omega$, such that the support of $f_j$ is contained in $K$ for all $j=0,1,2,\dots$,
(ii) for every $\epsilon > 0$ and every nonnegative interger $N$ there exists a nonnegative integer $m$ such that $\left| \left| f_j - f \right| \right|_N < \epsilon$ for all $j \geq m$.
Now, since $V \cap \mathcal{D}_K \in \tau_k$, there exists $\epsilon > 0$ and a nonnegative integer $N$ such that the set \begin{equation} B = \{ g \in \mathcal{D}_K : \left| \left| g - f \right| \right|_N < \epsilon \} \end{equation} is contained in $V \cap \mathcal{D}_K$. Then, if $m$ is the nonnegatve integer whose existence is stated in (ii), we conclude that $f_j \in V$ for all $j \geq m$. So there is no sequence $(f_j)$ in $A$ converging to $f$.
QED
NOTE. From NOTE (2) in my answer to the post Topology of the space $\mathcal{D}(\Omega)$ of test functions, we know that $A$ is dense in $\mathcal{D}(\Omega)$, and since $A$ is sequentially closed, we can conclude that $A$ is an example of a dense subset of $\mathcal{D}(\Omega)$ which is not sequentially dense in $\mathcal{D}(\Omega)$.
Yes, it is well-known that
Let $X$ be a sequential space. Then $X$ is countably compact iff $X$ is sequentially compact.
To be clear about definitions: $X$ is countably compact iff every countable open cover of $X$ has a finite subbcover. I'll use the convenient equivalence that $X$ is countably compact iff every infinite subset $A$ of $X$ has an $\omega$-accumulation point $p \in X$, i.e. every neighbourhood $U$ of $p$ has $U \cap A$ infinite. I wrote the proof of that equivalence (also free of separation axiom assumptions) here, in case this was unknown to you.
And $X$ is sequentially compact iff every sequence $(x_n)_n$ in $X$ has a convergent subsequence.
$X$ is sequential if for all subsets $A$ of $X$: $A$ sequentially closed iff $A$ is closed.
Proof (following the standard reference Engelking, General Topology 2nd ed. Theorem 3.10.31 (due to Franklin (1965), generalising this from metric spaces where it had been shown by Hausdorff in 1914). I'll be extra pedantic in my version of its proof to make sure I make no hidden separation axiom assumptions (Engelking includes Hausdorff in his definition of countable compactness and sequential compactness, so there are subtle instances in his proofs where he might use them, which I want to avoid for maximal generality)
If $X$ is sequentially compact, let $A$ be an infinite subset of $X$. Let $(a_n)_n$ be sequence of points from $A$ so that $a_n \neq a_m$ whenever $n \neq m$ (i.e. find an injection $\Bbb N \to A$). By sequential compactness, there is some $p \in X$ and some subsequence $(a_{n_k})_k$ of $(a_n)_n$ such that $a_{n_k} \to p$ as $k \to \infty$. Then if $U$ is any neighbourhood of $p$, there exists some $N$ so that for all $k \ge N$ we have $a_{n_k} \in U$. It follows that $\{a_{n_k} \mid k \ge N\} \subseteq U \cap A$ and so $U \cap A$ is infinite (by the injectivity). Hence $p$ is an $\omega$-limit point of $A$ and by the mentioned equivalence, $X$ is countably compact. This shows one implication.
Let $X$ be countably compact. Let $(x_n)_n$ be a sequence in $X$. We want to show it has a convergent subsequence.
We can assume WLOG that $n \neq m \to x_n \neq x_m$ (if you believe this read on, if you want my argument for it, reveal spoiler)
For $x \in X$ we can define $N_x=\{n \in \Bbb N\mid x_n = x\}$. If some $N_x$ is infinite, then $N_x$ defines a subsequence of $(x_n)_n$ that is constant with value $x$ and in any space this converges to $x$, and we'd be done. So assume all $N_x$ are empty (probably most of them are) or finite; picking one index from each non-empty one (yes I believe in AC) we get a subsequence as claimed, and as a subsequence of a subsequence is still a subsequence, we only have to consider that situation. We resume the proof under this injectivity assumption which is a handy technicality.
Define $A = \{x_n\mid n \in \Bbb N\}$ which is an infinite set. As $X$ is countably compact, $A$ has an $\omega$-accumulation point $p$. It's clear that the $A\setminus \{p\}$ is not closed, as $p$ is in its closure but not in the set. Because $X$ is sequential (finally we use it!), $A\setminus \{p\}$ is not sequentially closed, i.e. there is a sequence $(y_n)_n$ in $A \setminus \{p\}$ and some point $q \notin (A\setminus \{p\})$ so that $y_n \to q$.
By re-ordering this sequence we find a subsequence of $(x_n)_n$ converging to $q$. (details follow as before, skip if you see it already).
First note as all $y_n$ come from $A\setminus \{p\}$ and the sequence is injective, for every $n$ there is a unique $r(n) \in \Bbb N$ so that $y_n = x_{r(n)}$. Another folklore fact: if $h: \Bbb N \to \Bbb N$ is a bijection then $(y_{h(n)})_n \to q (n \to \infty)$ as well (any re-ordering of a convergent sequence has the same limit): let $U$ be any neighbourhood of $q$. There is an $N_1 \in \Bbb N$ so that $n > N_1$ implies $y_n \in U$. Then $h^{-1}[\{1,\ldots, N_1\}] \subseteq \{1,\ldots, N\}$ for some $N \in \Bbb N$, as $h^{-1}$ preserve finiteness. Then $h(n) > N$ implies $n > N_1$ so $q \in U$, and as $U$ was arbitrary, $y_{h(n)} \to q$ as $n \to \infty$. Then enumerate $r[\Bbb N]$ as $n_1< n_2< n_3 <n_4< \ldots$ and note that $x_{n_k} = (y_{r^{-1}(k)})$ converges to $q$ (as $k \to \infty$) as a re-ordering of $(y_n)_n$. So we have the desired convergent subsequence.
QED.
Best Answer
Consider the following operation on a subset $A$ of a space $X$, defining a new subset of $X$: $$\mbox{s-cl}(A) = \{ x \in X \mid \mbox{ there exists a sequence } (x_n)_n \mbox{ from } A \mbox{ such that } x_n \rightarrow x \}\mbox{.}$$ This set, the sequential closure of $A$, contains $A$ (take constant sequences) and in all spaces $X$ it will be a subset of the $\mbox{cl}(A)$, the closure of $A$ in $X$.
We can define $\mbox{s-cl}^{0}(A) = A$ and for ordinals $\alpha > 0$ we define $\mbox{s-cl}^\alpha(A) = \mbox{s-cl}(\cup_{\beta < \alpha} \mbox{s-cl}^\beta(A))$, the so-called iterated sequential closure.
A space is Fréchet-Urysohn when $\mbox{s-cl}(A) = \mbox{cl}(A)$ for all subsets $A$ of $X$, so the first iteration of the sequential closure is the closure.
A space is sequential if some iteration $\mbox{s-cl}^\alpha(A)$ equals the $\mbox{cl}(A)$, for all subsets $A$.
So basically by taking sequence limits we can reach all points of the closure eventually in a sequential space, but in a Fréchet-Urysohn space we are done after one step already.
For more on the differences and the "canonical" example of a sequential non-Fréchet-Urysohn space (the Arens space), see this nice topology blog, and the links therein.