Is this proof correct?
Let $\alpha$ an ordinal and let $\tau_\leq$ the order topology on $[0,\alpha)$. Let's prove that $[0,\alpha)$ is $T_4$, i.e., it separates closed sets with open sets.
Let $A,B\subseteq[0,\alpha)$ closed sets. I will define by transfinite induction two disjoint nondecreasing family of open sets. First: $$U_0=\phi,\quad V_0=\phi$$ If $\beta+1\leq\alpha$ is not a limit ordinal: note that $\{\beta+1\}=(\beta,\beta+2)$ is open. If $\beta+1\in A$, define $U_{\beta+1}=U_\beta\cup\{\beta+1\}$ and $V_{\beta+1}=V_\beta$. If $\beta+1\in B$, do the symmetric thing. If $\beta+1\not\in A\cup B$, do nothing.
If $\beta\leq\alpha$ is a limit ordinal, define $$U_\beta=\bigcup_{\beta'<\beta}U_{\beta'},\quad V_\beta=\bigcup_{\beta'<\beta}V_{\beta'}$$
Then, the families are clearly nondecreasing and consist of open sets. Suppose they are not disjoint and choose the first $\beta$ for which $U_\beta\cap V_\beta=\phi$. Clearly, $\beta$ cannot be a non limit ordinal. If $\beta$ is limit and $\beta'<\beta$ is such that $x\in U_{\beta'}$, then $x\not\in V_{\beta'}$. Since $\{U_\nu\}$ is nondecreasing, this works for all $\beta''\in[\beta',\beta)$ and, therefore, $x\not\in V_\beta$. So, $U_\alpha\cap U_\beta=\phi$.