Solution: Let $\mathcal{U} = \left\{ U_\alpha \right\}_{\alpha \in I}$ be an open cover of $X$. We need to show that $\mathcal{U} $ has a finite subcover. Let $U\in \mathcal{U} $. Since $U$ is open, so it either $X$ or $\emptyset$ or $U^c = X\setminus U$ is finite. Since $U$ is non-empty, it can be either $X$ or the complement is finite. If it is $X$, $U$ itself covers $X$ and we are done. Let $U\neq X$, so $U^c$ is finite. Let $U^c = \left\{ x_1,x_2,\ldots, x_k \right\} $. We choose $U_i\in \mathcal{U} $ such that $x_i\in U_i$, for $i=1,2,\ldots,k$. Note tha \[ X = U \cup U_1 \cup U_2 \cup \cdots \cup U_k \] Thus we obtained a finite subcover of $\mathcal{U} $, and hence, $X$ is compact.