Topological Completeness is not Hereditary

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

Let $H \subseteq S$ be a subset of $S$.

Let $\left({H, \tau_H}\right)$ be the topological subspace of $T$ induced by $H$.

Then it is not necessarily the case that $\left({H, \tau_H}\right)$ is also topologically complete.

That is, topological completeness is not hereditary.

Proof
Let $\left({\R, d}\right)$ be the real number line under the Euclidean metric.

Let $\left({\Q, d}\right)$ be the rational number space under the Euclidean metric.

We have that $\Q \subset \R$ by definition.

Let $T = \left({\R, \tau_d}\right)$ be the topological space induced on $\R$ by $d$.

By Real Number Line is Complete Metric Space, $\left({\R, d}\right)$ is a complete metric space.

Thus, by definition, $T$ is topologically complete.

Let $T' = \left({\Q, \tau_d}\right)$ be the topological space induced on $\Q$ by $d$.

By Rational Number Space is not Complete Metric Space, $\left({\Q, d}\right)$ is not a complete metric space.

Thus, by definition, $T'$ is not topologically complete.

Hence the result.