Compact in Subspace is Compact in Topological Space

Theorem
Let $T=\left(S,\tau\right)$ be a topological space.

Let $K\subseteq S$ be a subset.

Let $\tau_{K}$ be the subspace topology on $K$.

Let $T'=\left(K,\tau_K\right)$ 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 $\{W_i\}_{i \in J}$ be an open cover of $H$ in $T$.

Since $H\subseteq K$, $H\subseteq \bigcup_{i\in J} W_i \cap K$.

Since $W_i \cap K \in \tau_K$ for any $i\in J$, $\{W_i \cap K \}_{i \in J}$ is an open cover of $H$.

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

Since $H\subseteq \bigcup_{i=1}^{r} \left( W_i\cap K \right)$, $H\subseteq \bigcup_{i=1}^{r} W_i$.

Thus $\{W_i \}_{i =1}^{r}$ is an open cover of $H$ in $T$. Therefore $H$ is compact in $T$.