Topological Subspace is Topological Space/Proof 2

Theorem
Let $\left({X, \tau}\right)$ be a topological space.

Let $H \subseteq X$ be a non-empty subset of $X$.

Let $\tau_H = \left\{{U \cap H: U \in \tau}\right\}$ be the subspace topology on $H$.

Then the topological subspace $\left({H, \tau_H}\right)$ is a topological space.

Proof
Follows directly from Subspace Topology is Initial Topology with respect to Inclusion Mapping.

Also see

 * Topological Subspace