Solution: Take an open cover of $X / {\sim }$, say $\mathcal{U} = \left\{ U_\alpha : \alpha \in \mathcal{I} \right\} $, where $\mathcal{I} $ is an index set. We will prove that it has a finite subcover. Let $q : X \to X / \mathord{\sim }$ is the quotient map. For each $\alpha \in \mathcal{I} $, the set $U_\alpha $ is open in $X / \mathord{\sim }$, which means $q^{-1} \left( U_\alpha \right) $ must be open in $X$. Since $\mathcal{U} $ is an open cover of $X / \mathord{\sim }$, so $\left\{ V_\alpha = q^{-1} \left( U_\alpha \right) \right\} $ must be an open cover of $X$. As $X$ is compact, there will a finite subcover. Let $V_{\alpha _1}, V_{\alpha _2}, \dots, V_{\alpha _k}$ covers $X$. Then $U_{\alpha _1}, U_{\alpha _2}, \dots, U_{\alpha _k}$ will cover $X / \mathord{\sim }$, as $q$ is a surjective map. Thus, $X / \mathord{\sim }$ is compact.