What is the proof that first countable is sufficient to say that sequentially closed implies closed I've seen this statement quoted many times but I've been unable to find a proof of the statement.
I've been attempting to prove it myself with the following method:
Let $X$ be a topological space, with $A\subseteq X$ sequentially closed
Assume $A$ not closed. So there exists an $x\in X\setminus A$ such that every open set containing $x$ intersects with $A$.
So let $B_n$ be the countable neighbourhood basis at $x$.
And then define a sequence $x_n$ such that $x_n\in B_n\cap A$.
But this is where I'm stuck. I can't figure out how to show that $x_n$ approaches $x$.
Clearly every open set $C$ containing $x$ also contains a $B_i$ in the basis. and so it contains the corresponding $x_i$, but it seems that I would need some sort of nesting property on the neighbourhood basis (so that it looks like the open balls in $\mathbb R^n$ ) to show that the entirety of the tail of the sequence is contained in it.
So I went on in an attempt to prove some kind of nesting statement (which required $T^1$) but then it didn't even help prove the statement anyway because what I chose to prove wasn't powerful enough.
I'm realising that I will need to manipulate the basis $B_i$ in some way to approximate the nesting, because obviously leaving an arbitrary order for the basis won't force $x_n$ to approach $x$ (for example, in $\mathbb R$. let $B_n$ be the ball of radius $\frac1n$ when $n$ is even, and $B_n$ is the ball of radius 1 when $n$ is odd.)
Any advice for how to proceed or simply an example of the proof would be greatly appreciated.
 A: You're right about the nesting: you can force that with just countably many neighbourhoods: if $(B_n), n=0,1,2,\ldots$ is a local base at $x$, then define for every $n=0,1,2,\ldots$: $O_n = \bigcap_{i=0}^n B_i$, which are open neighbourhoods of $x$ as we have finite intersections of open sets. And when $O$ is an open set containing $x$, we have some $m$ with $x \in B_m \subseteq O$ and then $x \in O_m \subseteq B_m \subseteq O$ as well, so the $O_n$ also form a local base at $x$, but additionally satisfy $O_{n+1} \subseteq O_n$ for all $n$ (we take the intersection with one more set so we get a smaller set). Now picking the $x_n$ as you do, for the $O_n$ instead of the $B_n$, will work. No $T_1$ is required, as you see.
Many topologists just assume this basic fact known, and say when we have a first countable space, "let $(B_n)$ be a decreasing (nested) local base at $x$". We can always do this, as we saw, and it has the added benefit that however we choose $x_n \in B_n$ we always have a sequence converging to $x$. This is sort of modelled after the standard local base $B(x,\frac{1}{n})$ for a metric space, e.g. Often the natural local countable base is already nested.
