@AdamRubinson Okay, I can try to explain better if you explain what is not clear.

@RyszardSzwarc So I assume that $x_0$ is not an isolated point and then I discovered that $\operatorname{cl} Y\setminus\{x_0\}$ is not closed.

So by sequentiality there exists a sequence $(x_n)_{n\in\Bbb N}$ in $\operatorname{cl} Y\setminus\{x_0\}$ converging to some $y\notin\operatorname{cl}Y\setminus\{x_0\}$, right?

Now if $x_n$ is for each $n\in\Bbb N$ in $\operatorname{cl}Y\setminus\{x_0\}$ then $y$ must be in $\operatorname{cl}\big(\operatorname{cl}Y\setminus\{x_0\}\big)$, right?

But if $\operatorname{cl}\big(\operatorname{cl}Y\setminus\{x_0\}\big)$ is equal to $\operatorname{cl} Y$ and if $y$ must not be in $\operatorname{cl}Y\setminus\{x_0\}$ then necessarly $y$ must be equal to $x_0$.

@RyszardSzwarc Sorry, could you edit the latex?

I think the exercise is not translated correctly from the original (in Spanish). What the exercise says (8(a)) is to show that every closed subset of a sequential set is also sequential, and that Avery Urysoh-Frechet space is sequential. As for (8(b)), it is not written what is Proposition 3.42.

@OliverDíaz Proposition 3.42 asserts that $f:X\to y$ is continuous at $x$ if and only if $x_n\to x$ implies $f(x_n)\to f(x)$ when $X$ is first-countable.

@RyszardSzwarc Okay, so is a sequential space a Fréchet-Uryshon space? Where is my confusion?

@OliverDíaz Yes, right.

@RyszardSzwarc Okay, anyway thanks very much for your help!

@OliverDíaz I understand and I proved that a UF space is sequential: the problem is that if the proposition $3.41$ holds for sequential is ture that a sequential space if UF? This is my problem. Forgive my confusion.

@OliverDíaz So what can you say about?

@OliverDíaz I changed the image adding the proposition $3.42$ and the definition of UF. So do you want I traslate what is written there?

@AntonioMariaDiMauro: I completed my edits. I hope this is what you needed.

@OliverDíaz I saw: your translation seems good. So do you think that the result of the exercise $8$.b is true? This I would like to know: indeed as I explain it seems very stranger, that's all.

@user264745 I saw but unfortunately your liked question does not seem help: can you expalin better, please?

@OliverDíaz Anyway thanks very much for relevant edit!

The problem is to show that FU implies sequential. Not the other way around. In fact, the posting you are referring to (I have not read it carefully) tries to give and example of a space that is sequential but not FU, and that does not contradict the stament FU implies sequential.

@OliverDíaz Okay, I understood: however if the proposition $3.41$ holds for sequential then does this implies that a sequential space is FU? exactly this is my problem. If you like to follow I can prove that FU implies sequential provided you believes this is not clear. So did you understand what is my problem?

I see your problem now. It does seem that part (b) of the problem concerning 3.41 is incorrect.

@OliverDíaz Exactly but it seems that Engelking here use this: it is very stranger.

Hi, I am here!

It could be that there is no standard definition of "sequential convergence" and some authors add some additional and inconspicuous assumptions and some others not. But from what the textbook uses as a definition, it does seem that problem (b) regarding 3.41 is not correct, otherwise, sequential would be the same as FU. I have not done point set topology in quite a (25 years) in depth. I will check my old textbooks (Kelley, Dugundji, Willard)

Okay, so I will attend you, okay?

Anyway this morning I check Willard (General Topology) and it seems it treated only sequential compact spaces.