Subsequence convergence in hausdorff topological space

It’s false as stated.

Let $mathscr{U}$ be a free ultrafilter on $Bbb N$, and let $X=Bbb Ncup{mathscr{U}}$. Points of $Bbb N$ are isolated, and the sets ${mathscr{U}}cup U$ for $Uinmathscr{U}$ form a nbhd base at $mathscr{U}$. In other words, the topology on $X$ is

$$tau=wp(Bbb N)cupbig{{mathscr{U}}cup U:Uinmathscr{U}big};.$$

Let $sigma=langle n:ninBbb Nrangle$; every open nbhd of $mathscr{U}$ has nonempty intersection with $Bbb N$, but no subsequence of $sigma$ converges to $mathscr{U}$.

To see this, suppose that $langle n_k:kinBbb Nrangle$ is a subsequence of $sigma$ that converges to $mathscr{U}$, and let $A={n_k:kinBbb N}$. If $Anotinmathscr{U}$, then $Bbb Nsetminus Ainmathscr{U}$, and ${mathscr{U}}cup(Bbb Nsetminus A)$ is an open nbhd of $mathscr{U}$ disjoint from $A$; this is impossible, $langle n_k:kinBbb Nrangle$ converges to $mathscr{U}$, so $Ainmathscr{U}$. Let $A_0={n_k:ktext{ is even}}$ and $A_1={n_k:ktext{ is odd}}$. Exactly one of $A_0$ and $A_1$ belongs to the ultrafilter $mathscr{U}$; without loss of generality suppose that $A_0inmathscr{U}$. Then on the one hand $langle n_k:kin A_1rangle$ converges to $mathscr{U}$, since it’s a subsequence of $langle n_k:kinBbb Nrangle$, but on the other hand ${mathscr{U}}cup A_0$ is an open nbhd of $mathscr{U}$ disjoint from $A_1$. This contradiction shows that $sigma$ cannot in fact have a subsequence converging to $mathscr{U}$.

The problem is that $X$ is not a Fréchet-Uryson space: the topology at the point $mathscr{U}$ is too far away from being first countable.

You need something like first countability to actually get a subsequence. Otherwise there may be too many open sets.

Example: Let $X$ be the space of functions ${0,1}^{mathbb{N}}to{0,1}$ equipped with the topology of pointwise convergence, i.e., the product topology ${0,1}^{{0,1}^{mathbb{N}}}$. Let $f_nin X$ be $f_n((x_m)_{minmathbb{N}})=x_n$. Then a diagonal argument (*) shows that $f_n$ has no convergent subsequences. However, $X$ is compact, so countably compact and hence every countable subset $A$ of $X$ has an accumulation point $x$ (i.e., every neighbourhood of $x$ contains infinitely many points of $A$).

(*) The diagonal argument: suppose it has a convergent subsequence $f_{n_k}to f$. That means $f_{n_k}(x)to f(x)$ for all $xin{0,1}^{mathbb{N}}$. Let $x$ be the sequence that is $1$ at $n_{2ell}$ for all $ell$ and $0$ otherwise. Then $f_{n_k}(x)$ along odd and even $k$ are different constant sequences, contradicting $f_{n_k}(x)to f(x)$.