Restriction of Homeomorphism is Homeomorphism

Theorem
Let $T_1 = (S_1, \tau_1)$, $T_2 = (S_2, \tau_2)$ be topological spaces.

Let $f: S_1 \to S_2$ be a homeomorphism between $T_1$ and $T_2$.

Let $S$ be a subset of $S_1$.

Let $f \restriction_{S \times f\left[{S}\right]} : S \to f\left[{S}\right]$ be the restriction of $f$ to $S \times f\left[{S}\right]$.

Let $S$ and $f\left[{S}\right]$ bear the subspace topology.

Then $f \restriction_{S \times f\left[{S}\right]}$ is a homeomorphism between $S$ and $f\left[{S}\right]$.

Proof
By Restriction of Continuous Mapping is Continuous, $f \restriction_{S \times f\left[{S}\right]}$ is continuous.

By Restriction of Inverse is Inverse of Restriction, $(f \restriction_{S \times f\left[{S}\right]})^{-1}$ is well-defined and equal to $f^{-1} \restriction_{f\left[{S}\right] \times S}$.

By Restriction of Continuous Mapping is Continuous, $f^{-1} \restriction_{f\left[{S}\right] \times S}$ is continuous.

Since $(f \restriction_{S \times f\left[{S}\right]})^{-1} = f^{-1} \restriction_{f\left[{S}\right] \times S}$, $(f \restriction_{S \times f\left[{S}\right]})^{-1}$ is continuous.

Hence, $f \restriction_{S \times f\left[{S}\right]}$ is a homeomorphism.