Is there a non-Hausdorff space being a union of a countable increasing chain of "nice" spaces? For a chain $X_0 \subseteq X_1 \subseteq X_2 ...$ of topological spaces (i.e the topology on $X_n$ coincides with the subspace topology induced by $X_m$ whenever $m > n$) we consider the space with the underlying set $X = \bigcup_{n\in \mathbb{N}} X_n$ and the topology such that $U\subseteq X$ is open if and only if for every $n\in \mathbb{N}$ the set $U \cap X_n$ is open in $X_n$.
The question is whether there exists a chain $X_0 \subseteq X_1 \subseteq ...$ of "nice" spaces (nice being $T_3$/$T_4$, or at least Hausdorff), such that the space $X$ defined as above is not even Hausdorff.
The genesis of my problem is such: in the short article "A Regular Space on which every Continuous Real-Valued Function is Constant" by T. E. Gantner the author seemingly concludes that a space $X$ as above is regular based solely on the fact that $X_n$s are regular. Perhaps he uses the fact, that each $X_n$ is closed in $X_{n+1}$ (in the proof in the article it holds), but it is not clear and this fact does not look helpful either way.
My thoughts:

*

*If each $X_n$ is open in $X_{n+1}$, then the resulting space $X$ is always Hausdorff as long as all $X_n$s are Hausdorff.

Proof: It is easy to see that in fact $X_n$ is open in $X_{n+k}$ for every $n,k\in \mathbb{N}$ (say, by induction). Let $x,y \in X$ and let $m\in \mathbb{N}$ such that $x,y \in X_m$. Since $X_m$ is Hausdorff we can choose disjoint open $U_x, U_y\subseteq X_m$ such that $x\in U_x, y\in U_y.$ For every $n<m$ sets $U_x\cap X_n, U_y\cap X_n$ are open just by the definition of subspace topology, and for every $n>m$ sets $U_x\cap X_n = U_x, U_y\cap X_n = U_y$ are open by the assumption. So $U_x, U_y$ are in fact open in $X$ and they are disjoint open neighbourhoods of $x,y$, respectively. $\square$
That said, one can find a chain of "very nice" spaces $X_n$ such that $X$ is barely Hausdorff (i.e. is not $T_3$), namely let $X_n = \mathbb{R} \setminus \{\frac{1}{k} : k > n\}$, then $X$ is the very well known example of Hausdorff non-$T_3$ space (it is not possible to separate closed set $\{\frac{1}{k} : k > 0\}$ and point 0 with open sets). So the above cannot be strengthened, at least not without being very demanding about $X_n$s.


*If we do not know anything about how $X_n$ lies in $X_{n+1}$ (or we know that they are closed subspaces - as I do not know how to reasonably use that assumption), then I can mainly say that $X$ is $T_1$ if only each of the $X_n$s is $T_1$ (and that for $T_2/T_3/T_4$ spaces $X_n$ space $X$ is $T_2/T_3/T_4$ with "open" in respective definition replaced with $\text{"$G_\delta$"}$, so it seems pretty weak).

The natural reasoning here for any $T_k$ is similiar, so I will show it for $T_1/T_2$ (simultaneously): for $x,y \in X$ take $m\in \mathbb{N}$ such that $x,y \in X_m$. Take $U^m_x \ni x$ (and $U^m_y \ni y$ for $T_2$) - open subset of $X_m$ as in the definition of $T_1(T_2)$ space. As before for $n< m$ set $U^m_x\cap X_n (U^m_y\cap X_n)$ is open in $X_n$, so we move to $n>m$, first $n=m+1$. We want open set $U^{m+1}_x (U^{m+1}_y) \subseteq X_{m+1}$ such that $U^{m+1}_x\cap X_m = U^m_x$ (and $U^{m+1}_y \cap X_m = U^m_y$ for $T_2$). For $T_1$ we do not want anything else, as $y\notin U^{m+1}_x$ as required, and we can just continue this way, declaring $U = \bigcup_{n=m}^\infty U^n_x\subseteq X$, which is open in $X$ and satisfies $T_1$ requirement for points $x,y$. For $T_2$ we would like something similiar but, regrettably, we have no guarantee that $U^{m+1}_x \cap U^{m+1}_y$ is empty. Since $X_{m+1}$ is $T_2$ we can just choose disjoint open $V^{m+1}_x \ni x$ and $V^{m+1}_y \ni y$, but they do not need to have open intersections with $X_m$. It looks like the best thing for us to do is to switch sets $U^m_x, U^m_y$ for $U^{m+1}_x \cap V^{m+1}_y$ and $U^{m+1}_x \cap V^{m+1}_y$, respectively, which are disjoint, open in $X_{m+1}$ and have open intersections with $X_m$. But then we have to follow with this kind of switching infinitely many times and it is (a priori) possible that we end up with non-open $G_\delta$s, which is not enough for us.
 A: Let $Y$ be any non-completely normal space and let $A,B\subseteq Y$ be separated sets that cannot be separated by open sets.  Let $X=(Y\times\mathbb{N})\cup\{a,b\}$ and define a topology on $X$ by saying a subset $U\subseteq X$ is open if it satisfies the following conditions:

*

*$U\cap (Y\times\mathbb{N})$ is open in $Y\times\mathbb{N}$ (with respect to the product topology, where $\mathbb{N}$ is discrete).

*If $a\in U$ then $A\times\{n\}\subseteq U$ for all but finitely many $n$.

*If $b\in U$ then $B\times\{n\}\subseteq U$ for all but finitely many $n$.

This space is not Hausdorff: $a$ and $b$ cannot be separated by disjoint open sets, since that would give a separation of $A$ and $B$ by open subsets of $Y$.  However, note that if $U_n=(Y\setminus (A\cup B))\times (\mathbb{N}\setminus\{0,\dots n\})$, then $X$ is the colimit of the subspaces $X_n=X\setminus U_n$.  Moreover, each $X_n$ is "as nice as $Y$, $A$, and $B$": it's just a disjoint union of finitely many copies of $Y$, together with a copy of $A\times \mathbb{N}$ together with a "point at infinity" whose neighborhoods contain all but finitely many of the $A\times\{n\}$'s and similarly a copy of  $B\times\mathbb{N}$ together with a "point at infinity".  In particular, if $Y$ is Hausdorff, or regular, or completely regular, then so is each $X_n$, and if $Y$ and $A$ and $B$ are all normal, then so is each $X_n$.  Also, if $A$ and $B$ are closed in $Y$ (which is possible iff $Y$ is not normal), then each $X_n$ will additionally be closed in $X$.
(For an explicit example where $Y$ and $A$ and $B$ are all normal, you could take $Y$ to be the Tychonoff plank $(\omega+1)\times(\omega_1+1)$ with $A=\{\omega\}\times \omega_1$ and $B=\omega\times\{\omega_1\}$.)
These counterexamples are optimal in the following sense.  First, if $X$ is the colimit of an increasing sequence $(X_n)$ of closed normal subspaces, then $X$ is normal, as a consequence of the Tietze extension theorem (see Is the colimit of an expanding sequence of $T_4$ spaces $T_4$?).  Second, if $X$ is the colimit of an increasing sequence $(X_n)$ of completely normal subspaces, then $X$ is Hausdorff.  Indeed, let $a,b\in X$ be distinct points.  For any $n$ such that $X_n$ contains both $a$ and $b$, we can separate $a$ and $b$ by disjoint open subsets $U_n$ and $V_n$ of $X_n$.  Then $U_n$ and $V_n$ are separated, so we can separate them by disjoint open subsets $U_{n+1}$ and $V_{n+1}$ of $X_{n+1}$.  Continuing this recursively, we get increasing sequences $(U_m)_{m\geq n}$ and $(V_m)_{m\geq n}$ with each $U_m$ and $V_m$ disjoint open subsets of $X_m$.  The unions $U=\bigcup U_m$ and $V=\bigcup V_m$ will then be open subsets of $X$ that separate $a$ and $b$.
