Compact in Subspace is Compact in Topological Space

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

Let $K \subseteq S$ be a subset.

Let $\tau_K$ be the subspace topology on $K$.

Let $T' = \struct {K, \tau_K}$ be the topological subspace of $T$ determined by $K$.

Let $H \subseteq K$ be compact in $T'$.

Then $H$ is compact in $T$.

Proof
Suppose that $H$ is compact in $T'$.

Let $\set {W_i}_{i \in J}$ be an open cover of $H$ in $T$.

Then $\displaystyle H \subseteq \bigcup_{i \in J} W_i$.

Then:

Since $W_i \cap K \in \tau_K$, $\set {W_i \cap K}_{i \in J}$ is an open cover of $H$ in $T'$.

Since $H$ is compact in $T'$, $\set {W_i \cap K}_{i \in J}$ has some finite subcover $\set {W_i \cap K}_{i = 1}^r$.

Therefore:

So $\set {W_i}_{i = 1}^r$ is an open cover of $H$ in $T$, which is a finite subcover.

As $\set {W_i}_{i \in J}$ is arbitrary:

Any open cover of $H$ has a finite subcover in $T$.

So $H$ is compact in $T$.