$[0,\omega_1]\times [0,\omega_1)$ 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.
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?

Let $X=[0,\omega_1]\times [0,\omega_1) \text{, and show}$
$$\beta X=[0,\omega_1]\times [0,\omega_1].$$ Hint: Use the pressing down lemma to first prove that every continuous real-valued function on $[0,\omega_1)$ is eventually constant.  Then prove $X$ is $C^*$-embedded in $[0,\omega_1]\times [0,\omega_1]$.
The disjoint closed sets $H$ and $K$ in Brian M. Scott's answer clearly do not have disjoint closures in $[0,\omega_1]\times [0,\omega_1]$, so that $X$ cannot be normal. 
