[proofplan] We compare the two open-cover definitions directly. In one direction, a cover of $(K,\tau_K)$ by subspace-open sets can be represented by intersections $K \cap U$ with $U$ open in $X$, and these ambient open sets form an [open cover](/page/Open%20Cover) of $K$ in $X$. In the other direction, an ambient open cover of $K$ becomes a subspace-open cover after intersecting each member with $K$, and a finite subcover in the subspace lifts back to the original ambient cover. [/proofplan] [step:Turn a subspace-open cover into an ambient open cover] Assume first that $K$ is compact as a subset of $X$. Let $I$ be an index set, and let $(V_i)_{i \in I}$ be an open cover of $(K,\tau_K)$. Thus $V_i \in \tau_K$ for every $i \in I$, and $K \subset \bigcup_{i \in I} V_i$. By the definition of the [subspace topology](/page/Subspace%20Topology), for each $i \in I$ there exists a set $U_i \in \tau$ such that $V_i = K \cap U_i$. Then $(U_i)_{i \in I}$ is a cover of $K$ by open subsets of $X$: if $x \in K$, then $x \in V_i$ for some $i \in I$, hence $x \in K \cap U_i$, and therefore $x \in U_i$. [guided] Assume that $K$ is compact as a subset of $X$. We want to prove that $(K,\tau_K)$ is compact, so we begin with an arbitrary open cover of $(K,\tau_K)$. Let $I$ be an index set, and let $(V_i)_{i \in I}$ be a family such that $V_i \in \tau_K$ for every $i \in I$ and $K \subset \bigcup_{i \in I} V_i$. The obstacle is that compactness of $K$ as a subset of $X$ applies to covers by sets open in $X$, while the sets $V_i$ are only known to be open in the subspace topology on $K$. The definition of the subspace topology removes this obstacle: for each $i \in I$, since $V_i \in \tau_K$, there exists a set $U_i \in \tau$ such that $V_i = K \cap U_i$. We now verify that the family $(U_i)_{i \in I}$ covers $K$ in the ambient space $X$. Let $x \in K$. Since $(V_i)_{i \in I}$ covers $K$, there exists an index $i \in I$ such that $x \in V_i$. Using $V_i = K \cap U_i$, we get $x \in U_i$. Hence every $x \in K$ lies in at least one $U_i$, so $K \subset \bigcup_{i \in I} U_i$. Since each $U_i$ lies in $\tau$, this is an open cover of $K$ by subsets open in $X$. [/guided] [/step] [step:Extract a finite subcover and restrict it back to $K$] Since $K$ is compact as a subset of $X$, there exist finitely many indices $i_1,\dots,i_n \in I$ such that $K \subset \bigcup_{m=1}^{n} U_{i_m}$. Intersecting both sides with $K$ gives $K \subset \bigcup_{m=1}^{n} (K \cap U_{i_m})$. Because $K \cap U_{i_m} = V_{i_m}$ for every $m \in \{1,\dots,n\}$, we have $K \subset \bigcup_{m=1}^{n} V_{i_m}$. Thus every open cover of $(K,\tau_K)$ has a finite subcover, so $(K,\tau_K)$ is compact. [/step] [step:Turn an ambient open cover into a subspace-open cover] Conversely, assume that $(K,\tau_K)$ is compact. Let $I$ be an index set, and let $(U_i)_{i \in I}$ be a family of sets in $\tau$ such that $K \subset \bigcup_{i \in I} U_i$. For each $i \in I$, define $V_i := K \cap U_i$. Since $U_i \in \tau$, the definition of $\tau_K$ gives $V_i \in \tau_K$ for every $i \in I$. Moreover, if $x \in K$, then $x \in U_i$ for some $i \in I$, and therefore $x \in K \cap U_i = V_i$. Hence $K \subset \bigcup_{i \in I} V_i$. So $(V_i)_{i \in I}$ is an open cover of $(K,\tau_K)$. [/step] [step:Lift the finite subspace subcover to the original ambient cover] By compactness of $(K,\tau_K)$, there exist finitely many indices $i_1,\dots,i_n \in I$ such that \begin{align*} K \subset \bigcup_{m=1}^{n} V_{i_m}. \end{align*} Using $V_{i_m} = K \cap U_{i_m} \subset U_{i_m}$ for each $m \in \{1,\dots,n\}$, we obtain $K \subset \bigcup_{m=1}^{n} U_{i_m}$. Thus every cover of $K$ by open subsets of $X$ has a finite subcover. Therefore $K$ is compact as a subset of $X$. Combining the two directions proves the equivalence. [/step]