Tychonoff plank

Remarks

The sets $A = \set{\omega_1} \times [0, \omega)$ and $B = [0, \omega_1) \times \set{\omega}$ are disjoint and closed in $T_\infty$. Let $U$ be an open set of $T_\infty$ containing $A$. For each $n < \omega$ there is $\alpha_n < \omega_1$ with $(\alpha_n, \omega_1] \times \set{n} \subseteq U$. The supremum $\alpha$ of the countably many $\alpha_n$ is still countable, so $\alpha + 1 < \omega_1$ and $(\alpha + 1, n) \in U$ for every $n$. Any open set $V$ containing $(\alpha + 1, \omega) \in B$ contains $(\alpha + 1, n)$ for all large $n$, so $V$ meets $U$. Hence $A$ and $B$ cannot be separated, and $T_\infty$ is not normal. Since $T_\infty$ is a subspace of the compact Hausdorff space $T$, it is a Tychonoff space; so an open subspace of a normal space need not be normal.