Sierpinski space is normal and not completely regular, so the $T_1$ hypothesis in the Urysohn corollary is not decoration

Sierpinski space $S=\{a,b\}$, $a\mathrm{\ne }b$, with ${\mathcal{T}}_{\mathrm{S}\mathrm{i}\mathrm{e}\mathrm{r}}=\{\mathrm{\varnothing },\{b\},S\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) is normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly) but not completely regular (Completely regular spaces and Tychonoff ($T_{3\frac{1}{2}}$) spaces), by FALSE: Every normal space is completely regular. The refutation there in fact shows $S$ is not even regular — the only open set containing $a$ is $S$ itself — and complete regularity then fails because it implies regularity (Every completely regular space is regular, and every Tychonoff space is $T_3$). And $S$ is not $T_1$: the singleton $\{b\}$ is not closed in $S$.

This is exactly why Under dependent choice a normal $T_1$ space is completely regular, so $T_4\Rightarrow T_{3\frac{1}{2}}$, and together with the implications already proved this is the whole classical chain proves its arrow for a normal $T_1$ space and not for a bare normal space: $S$ is a normal space for which the corollary's conclusion — complete regularity — genuinely fails, and the one hypothesis it lacks is $T_1$.

• This is not a new computation. Every fact used above is proved in FALSE: Every normal space is completely regular; this item only reads that refutation as the positive example it also is, and connects it to the hypothesis of Under dependent choice a normal $T_1$ space is completely regular, so $T_4\Rightarrow T_{3\frac{1}{2}}$, and together with the implications already proved this is the whole classical chain by name.