Topological Completeness is not Hereditary

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

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

Let $\struct {H, \tau_H}$ be the topological subspace of $T$ induced by $H$.

Then it is not necessarily the case that $\struct {H, \tau_H}$ is also topologically complete.

That is, topological completeness is not hereditary.

Proof
Let $\struct {\R, d}$ denote the real number line under the Euclidean metric.

Let $\struct {\Q, d}$ denote the rational number space under the Euclidean metric.

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

Let $T = \struct {\R, \tau_d}$ be the topological space induced on $\R$ by $d$.

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

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

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

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

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

Hence the result.