T1 Property is Hereditary

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space which is a $T_1$ (Fréchet) space.

Let $T_H = \left({H, \tau_H}\right)$, where $\varnothing \subset H \subseteq S$, be a subspace of $T$.

Then $T_H$ is a $T_1$ (Fréchet) space.

Proof
Let $T$ be a $T_1$ (Fréchet) space.

That is:
 * $\forall x, y \in S$ such that $x \ne y$, both:
 * $\exists U \in \tau: x \in U, y \notin U$
 * and:
 * $\exists U \in \tau: y \in U, x \notin U$

We have that the set $\tau_H$ is defined as:
 * $\tau_H := \left\{{U \cap H: U \in \tau}\right\}$

We can assume that $\varnothing \subset H \subset S$, that is, $H \ne \varnothing$ and $H \ne S$, otherwise the result is trivial.

Let $x, y \in H$ such that $x \ne y$.

Then as $x, y \in S$ we have that:
 * $\exists U \in \tau: x \in U, y \notin U$

and:
 * $\exists U \in \tau: y \in U, x \notin U$

Then both:
 * $U \cap H \in \tau_H: x \in U \cap H, y \notin U \cap H$

and:
 * $U \cap H \in \tau_H: y \in U \cap H, x \notin U \cap H$

and so the $T_1$ axiom is satisfied.