The following is an exercise from Elementos de Topología General by Ángel Tamariz Mascarúa and Fidel Casarrubias Segura.
First a couple of definitions:
Definiton A topological space $X$ is FrechétUrysohn if for very $A\subset X$ and $x\in\overline{A}$, there is a sequence $(x_n:n\in\mathbb{N})\subset A$ that converges to $x$.
Definition A topological space $X$ is said to be sequential if for any $Y\subset X$, $Y$ is not closed if and only if there exists $x\in X\setminus Y$ and a sequence $(x_n: n\in\Bbb N)\subset Y$ with converges to $x$.
Problem 7 and 8 in that textbook say:

7: Check that a first countable topological space is FrechétUryshohn, and that this property is inherited by any subspace of $X$.

8(a) Show that every closed subset of a sequential space is also sequential, and that every FréchetUryshohn space is sequential.

8(b) Check that propositions 3.41 and 3.42 (below) hold if the assumption first countable is replaced by sequential.
Here are the statement of the Propositions mentioned in the problem:
Proposition 3.41: If $X$ is first countable and $E\subset X$, then $x\in \overline{E}$ (closure of $E$) iff there is a sequence $(x_n:n\in\mathbb{N})\subset E$ that converges to $x$.
Proposition 3.42: Let $f:X\rightarrow Y$ be a map between topological spaces. If $X$ is first countable and $x\in X$, then $f$ is continuous at $x$ if and only if for every sequence $(x_n:n\in\mathbb{N})$ with $x_n\xrightarrow{n\rightarrow\infty}x$, $f(x_n)\xrightarrow{n\rightarrow\infty}y$ in $Y$.
So I am really confused about the preceding theorem becasue if it is was true then it seem to me it would prove that a sequential space is a FréchetUryshon space but as here showed this is false. So first of all to follow I put a direct reference of the text I mentioned hoping I understood bad Spanish text.
Anyway I tried to prove the statement and surprisingly it seems true. So to follow my proof attempt.
proof $\,8$.b.$1.\,\,$
So if $(x_n)_{n\in\Bbb N}$ is a sequence in $Y$ converging to $x_0$ then by convergence definition for any neighborhood $V$ of $x_0$ there exist $n_V\in\Bbb N$ such that
$$
x_n\in V
$$
for all $n\ge n_V$ but if $x_n$ is in $Y$ for all $n\in\Bbb N$ then this means that $V\cap Y$ is not empty, that is $x_0$ is adherent to $Y$.
Conversely let be $x_0\in\operatorname{cl}Y$ and we let use sequentiality to make a sequence on $Y$ converging to $x_0$. So if $x_0$ is an isolated point of $Y$ then trivially the position
$$
x_n:=x
$$
for all $n\in\Bbb N$ defines sequence in $Y$ converging to $x_0$ so we suppose that $x_0$ is an accumulation point for $Y$. Now if $x_0$ is an accumulation point for $Y$ then as here showed the identity
$$
\operatorname{cl}\big(\operatorname{cl}Y\setminus\{x_0\}\big)=\operatorname{cl} Y
$$
holds and thus we conclude that $\operatorname{cl}Y\setminus\{x_0\}$ is not closed so that by sequentiality there exists a sequence $(x_n)_{n\in\Bbb N}$ in $\operatorname{cl}Y\setminus\{x_0\}$ converging to $x\notin \operatorname{cl}Y\setminus\{x_0\}$. So if $x_n$ is in $\operatorname{cl}Y\setminus\{x_0\}$ for all $n\in\Bbb N$ then by the first implication $x$ must be in $\operatorname{cl}\big(\operatorname{cl}Y\setminus\{x_0\}\big)$ that is in $\operatorname{cl} Y$ and so $x$ must be equal to $x_0$ because the unique element of $\operatorname{cl} Y$ not in $\operatorname{cl}Y\setminus\{x_0\}$ is $x_0$.
So is the proposition $8$.b.$1$ true? if it is true then is the proof I gave correct? Could someone help me, please?
Best Answer
I write this to summarize my previous comments and Brian Scott's insightful comments.
Indeed, there must be a typo in the textbook to which you are referring. You may want to contact one of the authors to see if they keep tabs on errata. What I think problem 8 should have ask is to check whether propositions 3.41 and 3.42 hold when first countable is replaced by sequential.
Suppose $(X,\tau)$ is has the FrechétUrysohn property. If $A\subset X$ is not closed, then there exists $a\in \overline{A}\setminus A$. Consequently there exists a sequence $(x_n:n\in\mathbb{N})\subset A$ such that $x_n\xrightarrow{n\rightarrow\infty}a$. This implies that $(X,\tau)$ is sequential.
Prop 3.41 implies that every first countable space has the FrechétUrysohn property.
8b: