[proofplan] We prove compactness of $F$ by taking an arbitrary [open cover](/page/Open%20Cover) of $F$ in the [subspace topology](/page/Subspace%20Topology) and converting it into an open cover of the ambient [compact space](/page/Compact%20Space) $X$. Since $F$ is closed, the complement $X \setminus F$ is open, so adjoining it to the lifted cover gives an open cover of $X$. Compactness of $X$ gives a finite subcover, and removing the complement leaves finitely many original subspace-open sets covering $F$. [/proofplan] [step:Lift an arbitrary subspace-open cover of $F$ to ambient open sets] Let $\tau_F$ denote the subspace topology on $F$, so $\tau_F=\{F\cap U:U\in\tau\}$. Let $\mathcal V \subset \tau_F$ be an open cover of $F$. For each $V\in\mathcal V$, choose a set $U_V\in\tau$ such that $V=F\cap U_V$. Then the family $\mathcal W:=\{U_V:V\in\mathcal V\}\cup\{X\setminus F\}$ is a family of open subsets of $X$, because each $U_V\in\tau$ and $X\setminus F\in\tau$ since $F$ is closed. [/step] [step:Use compactness of $X$ to obtain finitely many lifted sets] The family $\mathcal W$ covers $X$. Indeed, if $x\in F$, then $\mathcal V$ covers $F$, so there exists $V\in\mathcal V$ such that $x\in V=F\cap U_V$, hence $x\in U_V$. If $x\in X\setminus F$, then $x$ is covered by $X\setminus F$. Since $X$ is compact, the open cover $\mathcal W$ has a finite subcover. Hence there exist $V_1,\dots,V_n\in\mathcal V$ such that \begin{align*} X=U_{V_1}\cup\dots\cup U_{V_n}\cup (X\setminus F) \end{align*} after possibly omitting $X\setminus F$ if it is not part of the chosen finite subcover. [guided] We need to use compactness of $X$, so we must build an open cover of $X$, not only of $F$. The subspace-open cover $\mathcal V$ covers only points of $F$, and its members need not themselves be open subsets of $X$. By the definition of the subspace topology, however, every $V\in\mathcal V$ has the form $V=F\cap U_V$ for some ambient [open set](/page/Open%20Set) $U_V\in\tau$. The points outside $F$ are not necessarily covered by the sets $U_V$, so we add the complement $X\setminus F$. This set is open in $X$ because $F$ is closed. Therefore \begin{align*} \mathcal W:=\{U_V:V\in\mathcal V\}\cup\{X\setminus F\} \end{align*} is an open cover of $X$. Now compactness of $X$ applies exactly: every open cover of $X$ has a finite subcover. Thus finitely many members of $\mathcal W$ cover $X$. The finitely many members coming from the lifted cover can be written as $U_{V_1},\dots,U_{V_n}$ for some $V_1,\dots,V_n\in\mathcal V$, and the finite subcover may also include $X\setminus F$. [/guided] [/step] [step:Restrict the finite subcover back to $F$] We claim that $V_1,\dots,V_n$ cover $F$. Let $x\in F$. Since the finite family from the previous step covers $X$, the point $x$ belongs to one of the sets $U_{V_1},\dots,U_{V_n}$ or to $X\setminus F$. The latter is impossible because $x\in F$. Hence $x\in U_{V_j}$ for some $j\in\{1,\dots,n\}$, and therefore \begin{align*} x\in F\cap U_{V_j}=V_j. \end{align*} Thus $\{V_1,\dots,V_n\}$ is a finite subcover of $\mathcal V$. Since $\mathcal V$ was an arbitrary open cover of $F$ in the subspace topology, $F$ is compact with the subspace topology inherited from $X$. [/step]