T1 is hereditary

This article gives the statement, and possibly proof, of a topological space property (i.e., T1 space) satisfying a topological space metaproperty (i.e., subspace-hereditary property of topological spaces)
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Statement

Any subspace of a T1 space, endowed with the subspace topology, is again a $T_1$-space.

Definitions used

T1 space

Further information: T1 space

A topological space $X$ is termed $T_1$ if it satisfies the following equivalent conditions:

1. For any two distinct points $x,y\in X$, there exists an open subset $U$ of $X$ such that $x\in U,y\notin U$
2. For every point $x\in X$, the singleton subset $\{x\}$ is closed in $X$
3. For every point $x\in X$, the intersection of all open subsets of $X$ containing $x$, is $\{x\}$

Subspace topology

Further information: Subspace topology

The subspace topology on a subset $A$ of $X$ is defined in the following equivalent ways:

1. A subset $U$ of $A$ is open in $A$ iff there exists an open subset $V$ of $X$ such that $V\cap A=U$.
2. A subset $C$ of $A$ is closed in $A$ iff there exists a closed subset $D$ of $X$ such that $D\cap A=C$.

Proof

Proof in terms of two-point definition of T1

Given: A $T_1$-space $X$, a subset $A$

To prove: $A$ is a $T_1$-space when endowed with the subspace topology

Proof: We need to show that if $x\ne y$ are both points of $A$, then there exists an open subset of $A$ containing $x$ and not containing $y$.

Since $x,y$ are distinct points of $A$, they are also distinct points of $X$. Since $X$ is a $T_1$-space, there exists an open subset $V$ of $X$ such that $x\in V$ and $y\notin V$. Now consider the set $U=V\cap A$. Then, by definition of subspace topology, $U$ is open in $A$. Further, since $x\in A$ and $x\in V$, we have $x\in U$. Since $y\notin V$, we have $y\notin U$. Thus, $U$ is an open subset of $X$ containing $x$ and not containing $y$.