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

1. 1.

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

2. 2.

   Every open cover of $Y$ (with open sets in $X$) has a finite subcover.

Proof. Suppose $Y$ is compact, and $\{U_{i}\}_{{i\in I}}$ is an arbitrary open 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 $Y$ with union $Y$. 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 cover of $Y$ 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. $\Box$

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