Exercise 2.5.3 ($T_{1}$-spaces, Hausdorff spaces and regular spaces are hereditary)

Let $X$ be a topological space, and let $Y\subseteq X$ be a subspace of $X$.

$T_1$ space

Proof. Suppose that $X$ is $T_1$. Pick an arbitrary singleton $\{x\},X\in Y$. Since $\{x\}$ is closed in $X$, $X\setminus \{x\}$ is open in $X$. By the definition of relative topology, $(X\setminus \{x\})\cap Y=(X\cap Y)\setminus \{x\}=Y\setminus \{x\}$ is relatively open in $Y$; thus, $\{x\}$ is closed in $Y$. Since our choice of $x\in Y$ was arbitrary, $Y$ must be a $T_1$-space. □

$T_2$ space

Proof. Suppose that $X$ is Hausdorff. Pick an arbitrary $x,y\in Y$. By the Hausdorff property, we can find disjoint open sets $U,V\subseteq X$ such that $x\in U$ and $y\in V$. Then sets $U\cap Y$ and $V\cap Y$ must be disjoint and relatively open sets in $Y$ containing $x$ and $y$ respectively. Thus, $Y$ must be Hausdorff, as well. □

regular space

Proof. Suppose that $X$ is regular. Let $E$ be a relatively closed subset of $Y$, and let $x\in Y\setminus E$. By Theorem 2.1, $E$ being a relatively closed subset of $Y$ implies that there is a closed subset $S$ of $X$ such that $E=S\cap Y$ (obviously $x\notin S$). By regularity, we can find two disjoint open sets $U$ and $V$ such that $S\subseteq U$ and $x\in V$. Intersecting these two sets with $Y$, we obtain two disjoint relatively open sets $U\cap Y$ and $V\cap Y$ in $Y$ containing $E$ and $x$ separately. □