Moore plane

Remarks

The $x$-axis $A = \set{(x, 0) : x \in \R}$ is closed, since every point with $y > 0$ has a disc neighbourhood missing $A$. The subspace $A$ is discrete, because a tangent disc neighbourhood of $(x, 0)$ meets $A$ only in $(x, 0)$. So every subset of $A$ is closed in $\Gamma$. The points with rational coordinates and $y > 0$ form a countable dense set, so $\Gamma$ is separable. By Jones' lemma, a separable normal space has no closed discrete subset $A$ with $2^{\abs{A}} > 2^{\aleph_0}$. Here $\abs{A} = 2^{\aleph_0}$, so $\Gamma$ is not normal. For example, the points of $A$ with rational $x$ and the points of $A$ with irrational $x$ are disjoint closed sets that cannot be separated by disjoint open sets. Since $\Gamma$ is Tychonoff, it is a subspace of a compact Hausdorff space, which is normal; so a subspace of a normal space need not be normal.