Non-Homeomorphic Sets may be Homeomorphic to Subsets of Each Other

Theorem
Let $T_1 = \left({S_1, \tau_1}\right)$ and $T_2 = \left({S_2, \tau_2}\right)$ be topological spaces.

Let $H_1 \subseteq S_1$ and $H_2 \subseteq S_2$.

Then it is possible for:
 * $(1): \quad T_1$ to be homeomorphic to $H_2$
 * $(2): \quad T_2$ to be homeomorphic to $H_1$

but:
 * $(3): \quad T_1$ and $T_2$ to not be homeomorphic.

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

Let $S_1 := \left[{0 \,.\,.\, 1}\right]$ be the closed unit interval.

Let $S_2 := \left({0 \,.\,.\, 1}\right)$ be the open unit interval.

Let $H_1 := \left({0 \,.\,.\, 1}\right)$ and $H_2 := \left[{\dfrac 1 4 \,.\,.\, \dfrac 3 4}\right]$.

Then:
 * $(1): \quad \left({S_1, \tau_d}\right)$ is homeomorphic to $\left({H_2, \tau_d}\right)$ by the mapping $f: S_1 \to H_2$ defined as $f \left({x}\right) = \dfrac x 2 + \dfrac 1 4$
 * $(2): \quad \left({S_2, \tau_d}\right)$ is trivially homeomorphic to $\left({H_1, \tau_d}\right)$

From Continuous Image of Compact Space is Compact, if $S_1$ and $S_2$ are homeomorphic then both must be compact.

From Closed Real Interval is Compact, $S_1$ is compact.

But from Open Real Interval is not Compact, $S_2$ is not compact.

Hence:
 * $(3): \quad \left({S_1, \tau_d}\right)$ and $\left({S_2, \tau_d}\right)$ are not homeomorphic.