Let $$X_1\subseteq X_2\subseteq\dots\subseteq X_n\subseteq\dots$$ be an expanding sequence of spaces. Write $X_\infty$ for the colimit of the sequence. i.e. $X_\infty=\bigcup X_n$ topologised so that a subset $U\subseteq X_\infty$ is open (closed) if and only if $U\cap X_n$ is open (closed) in $X_n$ for each $n$.
Assume that each $X_n$ is $T_4$ and that each inclusion $X_n\subseteq X_{n+1}$ is closed. Is it true that $X_\infty$ is $T_4$? If this is true, then is it required that the inclusions are closed?
Here $T_4$ will mean that disjoint closed sets have disjoint open neighbourhoods. Counterexamples are appreciated.
Note, if each $X_n$ is $T_1$ (points are closed), then $X_\infty$ is $T_1$ (you don't need the inclusions to be closed for this). Moreover if each $X_n$ is normal ($=T_1+T_4$), then a cute argument with the Tietze Extension theorem shows that $X_\infty$ is normal. If each $X_n$ is $T_2$, resp. $T_3$, then further assumptions are needed to guarantee that $X_\infty$ is $T_2$, resp. $T_3$.
Returning to the question let $C,D\subseteq X_\infty$ be disjoint closed sets and for each $n$ put $C_n=C\cap X_n$, $D_n=D\cap X_n$. If $U_n,V_n\subseteq X_n$ are disjoint open neighbourhoods of $C_n,D_n$ in $X_n$, then it would suffice to find disjoint open neighbourhoods $U_{n+1},V_{n+1}\subseteq X_{n+1}$ of $C_{n+1},D_{n+1}$ in $X_{n+1}$ such that $U_{n}\subseteq U_{n+1}$ and $V_n\subseteq V_{n+1}$. Inductively this gives expanding sequences of subspaces $\dots\subseteq U_n\subseteq U_{n+1}\subseteq\dots$ and $\dots\subseteq V_n\subseteq V_{n+1}\subseteq\dots$ and setting $U=\bigcup U_n$, $V=\bigcup V_n$ we get disjoint open neighbourhoods of $C,D$ in $X_\infty$.
The Tietze extension theorem does not require points to be closed, so your "cute argument" works just as well for $T_4$ spaces, assuming all the inclusions are closed. In detail, suppose $A,B\subseteq X_\infty$ are disjoint closed subsets. By Urysohn's lemma, there is a continuous function $f_1:X_1\to[0,1]$ such that $f_1$ is $0$ on $A\cap X_1$ and $1$ on $B\cap X_1$. We can then continuously extend $f_1$ to $X_1\cup((A\cup B)\cap X_2)$ by making it $0$ on all of $A\cap X_2$ and $1$ on all of $B\cap X_2$, and then by the Tietze extension theorem this extends continuously to a function $f_2$ on all of $X_2$. We can similarly extend $f_2$ to $f_3:X_3\to [0,1]$ that is $0$ on $A\cap X_3$ and $1$ on $B\cap X_3$ and so on. Gluing together all these functions gives a continuous $f:X_\infty\to [0,1]$ which is $0$ on $A$ and $1$ on $B$.
Here is a counterexample if the inclusions are not closed. Let $$X_\infty=(\mathbb{N}\times\{0,1\})\cup\{a\}$$ with the following topology. A set $U\subseteq X_\infty$ is open iff it satisfies the following conditions:
Note that $X_\infty$ is not $T_4$, since $\{a\}$ and $\mathbb{N}\times\{1\}$ are disjoint closed sets which do not have disjoint neighborhoods (any neighborhood of either must contain $(n,0)$ for all but finitely many $n$). Now let $$X_n=(\mathbb{N}\times\{0\})\cup\{a\}\cup (\{0,\dots,n-1\}\times\{1\})\subset X_\infty.$$ Clearly $X_\infty$ is the union of the $X_n$, and it is easy to see that it is in fact the colimit of them (if $U$ fails to satisfy one of the conditions above for being open, then that is detected in some $X_n$). I claim moreover that each $X_n$ is $T_4$. Indeed, $X_n$ is just the disjoint union of $(\{k\in\mathbb{N}:k\geq n\}\times\{0\})\cup\{a\}$ (which is a compact Hausdorff space, homeomorphic to $\mathbb{N}\cup\{\infty\}$) and $n$ 2-point spaces (namely, $\{(k,0),(k,1)\}$ for $k=0,\dots,n-1$). Any 2-point space is $T_4$, and a disjoint union of $T_4$ spaces is $T_4$, and thus each $X_n$ is $T_4$.
(You can get a similar example in which points are closed by replacing each point $(n,0)$ with an infinite set of points that accumulate at $(n,1)$. Or for a similar example that is more geometrical, let $X_n=[0,1]^2\setminus((0,1/n)\times\{0\})$ as a subspace of $\mathbb{R}^2$. The colimit is then $[0,1]^2$, but with its topology modified so that $(t,0)$ does not approach $(0,0)$ as $t$ approaches $0$. This is not normal since $\{(0,0)\}$ and $(0,1]\times\{0\}$ are disjoint closed sets that don't have disjoint neighborhoods.)