Proposition 12

Every subspace of a regular space is regular.

Let $A$ be a subspace of a regular space $X$. Since $X$ is Hausdorff, so is $A$ (no. 2); on the other hand, every neighbourhood of a point $x \in A$ with respect to $A$ is of the form $V \cap A$, where $V$ is a neighbourhood of $x$ in $X$. Since $X$ is regular there is a neighbourhood $W$ of $x$ in $X$ which is closed in $X$ and contained in $V$; $W \cap A$ is then a neighbourhood of $x$ in $A$, closed in $A$ and contained in $V \cap A$. Hence the result. Conversely: