$Y$ is compact if and only if every open cover of $Y$ has a finite subcover

Theorem.
Let $X$ be a topological space and $Y$ a subset of $X$. Then the followingstatements are equivalent.

1. $Y$ is compact as a subset of $X$.

2. Every open cover of $Y$ (with open sets in $X$) has a finitesubcover.

Proof.Suppose $Y$ is compact, and ${\{U_i\}}_{i\in I}$ is an arbitraryopen cover of $Y$, where $U_i$ are open sets in $X$. Then${\{U_i\cap Y\}}_{i\in I}$ is a collection of open sets in $Y$ withunion $Y$. Since $Y$ is compact, there is a finite subset $J\subset I$such that $Y={\cup }_{i\in J}(U_i\cap Y)$. Now$Y=({\cup }_{i\in J}{U}_i)\cap Y\subset {\cup }_{i\in J}{U}_i$, so${\{U_i\}}_{i\in J}$ is finite open cover of $Y$.

Conversely, suppose every open cover of $Y$ has a finite subcover,and ${\{U_i\}}_{i\in I}$ is an arbitrary collection of open sets(in $Y$) with union $Y$. By the definition of the subspace topology,each $U_i$ is of the form $U_i=V_i\cap Y$ for some open set$V_i$ in $X$. Now $U_i\subset V_i$, so ${\{V_i\}}_{i\in I}$ is a coverof $Y$ by open sets in $X$. By assumption, it has a finite subcover${\{V_i\}}_{i\in J}$. It follows that${\{U_i\}}_{i\in J}$ covers $Y$, and $Y$ is compact. $\mathrm{\square }$

The above proof follows the proof given in [1].