Proof that compactness of a subset can be tested with open sets of the ambient space

Proof

We translate covers in each direction with the definition of the subspace topology.

1. Let $X$ be a topological space and let $K \subseteq X$. Give $K$ the subspace topology, whose open sets are the sets $K \cap U$ with $U$ open in $X$.
2. By definition, $K$ is compact when $K$ with this topology is a compact space. That is, each open cover of $K$ by open sets of $K$ has a finite subcover.
3. Assume $K$ is compact. Let $(U_i)_{i \in I}$ be open sets of $X$ with $K \subseteq \bigcup_{i \in I} U_i$.
4. For each $i$, the set $V_i = K \cap U_i$ is open in $K$. Each $x \in K$ lies in some $U_i$, hence in $V_i$. So $(V_i)_{i \in I}$ is an open cover of the space $K$.
5. By compactness, there are $i_1, \ldots, i_n \in I$ with $K = V_{i_1} \cup \cdots \cup V_{i_n}$.
6. Since $V_{i_j} \subseteq U_{i_j}$, we get $K \subseteq U_{i_1} \cup \cdots \cup U_{i_n}$. This proves the first implication.
7. Now assume the cover condition for open sets of $X$. Let $(V_i)_{i \in I}$ be an open cover of the space $K$ by open sets of $K$.
8. For each $i$, choose an open set $U_i$ of $X$ with $V_i = K \cap U_i$. This is possible by step 1.
9. Then $K = \bigcup_i V_i \subseteq \bigcup_i U_i$. By the assumption, there are $i_1, \ldots, i_n \in I$ with $K \subseteq U_{i_1} \cup \cdots \cup U_{i_n}$.
10. Let $x \in K$. Then $x \in U_{i_j}$ for some $j$, and $x \in K$, so $x \in K \cap U_{i_j} = V_{i_j}$.
11. So $V_{i_1}, \ldots, V_{i_n}$ is a finite subcover of $(V_i)_{i \in I}$. Thus $K$ is a compact space, and $K$ is compact.