Suppose $X$ is a topological space equipped with the direct limit topology of the sequence $K_1\subset K_2\subset \cdots$ where each $K_n$ is compact Hausdorff. Thus a set $A\subset X$ is open [resp. closed] iff the intersection $A\cap K_n$ is open [resp. closed] in $K_n$, for each $n$. I am then trying to show that $X$ is itself regular.
Thus I have to show that $X$ is itself Hausdorff, and that if $C$ is a closed subset of $X$ and $x\in X-C$, then there are disjoint open sets $U,V\subset X$ such that $C\subset U$ and $x\in V$. It seems that the proofs of these two states will be similar, but I can't see how to prove these. Any hints?
First we show that the space $X$ is Hausdorff. Let $x,y$ be any distinct points of the space $X$. There exists a natural number $n$ such that $x,y\in K_n$. Since $K_n$ is Hausdorff compact space, it is Tychonoff and so there exists a continuous function $f_n: K_n\to\Bbb R$ such that $f_n(x)\ne f_n(y)$. Since each $K_m$ is a closed subspace of a normal space $K_{m+1}$, we can inductively extend $f_n$ to a continuous function $f_{n+1}$ from $K_{n+1}$ to $\Bbb R$, then to a continuous function $f_{n+2}$ from $K_{n+2}$ to $\Bbb R$ and so forth. Since $X=\bigcup K_m$, there is a function $f:X\to\Bbb R$ such that $f|K_m=f_m$ for each $m\ge n$. Since for each closed subset $F$ of $\Bbb R$ and each $m\ge n$, we have $f^{-1}(F)\cap K_m=f_m^{-1}(F)$, we see that the set $f^{-1}(F)$ is closed in $X$. So the map $f$ is continuous and $f(x)\ne f(y)$.
Note that the space $X$ is $k_\omega$. According to [FT], each Hausdorff $k_\omega$-space is normal. A bit stronger fact than regularity of a Hausdorff $k_\omega$-space is proved in our paper [BR], see Proposition 1 and definitions before it.
References
[BR] T. Banakh, A. Ravsky, On subgroups of saturated or totally bounded paratopological groups, Algebra and Discrete Mathematics, 4 (2003), 1-–20.
[FT] S.P. Franklin, B.V. Smith Thomas, A survey of $k_\omega$-spaces, Topology Proc. 2 (1977), 111–124.