T1 is hereditary

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

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
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 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 $$y$$.