Q:

$[0,\omega_1]\times [0,\omega_1)$ is not $T_4$.

With $T_4$ I mean a normal and $T_1$ space. I know this result is true, in fact there exists the following general theorem:
$X$ is paracompact iff $X\times \beta X$ is normal. Where $\beta X$ is the Stone-Cech compactification of $X$. The proof is given in "H. Tamamo, On paracompactness, Pacific J, Math. 10 (1960), 1043-1047". Here's a link 
(https://projecteuclid.org/download/pdf_1/euclid.pjm/1103038250).
And so $[0,\omega_1]\times [0,\omega_1)$ is not $T_4$ because $[0,\omega_1)$ is not paracompact. But there is another way of getting the result without making reference to that theorem? Maybe an easiear proof?

A:

Yes. Let $H=\{\omega_1\}\times[0,\omega_1)$, and let $K=\{\langle\alpha,\alpha\rangle:\alpha<\omega_1\}$; then $H$ and $K$ are disjoint closed subsets of $X=[0,\omega_1]\times[0,\omega_1)$ that cannot be separated by disjoint open sets.
Suppose that $U$ is an open nbhd of $K$. For each $\alpha<\omega_1$ there is an $f(\alpha)<\alpha$ such that 
$$\langle\alpha,\alpha\rangle\in\big(f(\alpha),\alpha\big]\times\big(f(\alpha),\alpha\big]\;,$$
since sets of this form are a local base at $\langle\alpha,\alpha\rangle$. Now $f:[0,\omega_1)\to[0,\omega_1)$ is a regressive (or pressing-down) function: for each $\alpha\in[0,\omega_1)$, $f(\alpha)<\alpha$. The pressing-down lemma (also known as Fodor’s lemma) implies that there are a $\beta\in[0,\omega_1)$ and an uncountable $S\subseteq[0,\omega_1)$ such that $f(\alpha)=\beta$ for each $\alpha\in S$. Thus,
$$U\supseteq\bigcup_{\alpha\in S}\Big(\big(f(\alpha),\alpha\big]\times\big(f(\alpha),\alpha\big]\Big)=(\beta,\omega_1)\times(\beta,\omega_1)\;.$$
Now let $V$ be any open nbhd of $H$, and let $\gamma\in(\beta,\omega_1)$; then $\langle\omega_1,\gamma\rangle\in H\subseteq V$, so there is an $\eta<\omega_1$ such that $\eta>\beta$ and $\{\omega_1\}\times(\eta,\omega_1]\subseteq V$. But then
$$\langle\eta+1,\gamma\rangle\in V\cap\big((\beta,\omega_1)\times(\beta,\omega_1)\big)=V\cap\bigcup_{\alpha\in S}\Big(\big(f(\alpha),\alpha\big]\times\big(f(\alpha),\alpha\big]\Big)\subseteq V\cap U\;,$$
so $V\cap U\ne\varnothing$, and $X$ isn’t $T_4$.

