How is every subset of the set of reals with the finite complement topology compact?

Let $X$ be any set endowed with the finite complement topology. Let $\mathcal{A}=\{A_i|i\in I\}$ be an open cover of $X$ (where $I$ is an arbitrary index set). Take any $A_0\in\mathcal{A}$, then $X\backslash A_0$ contains only a finite number of points $\{x_1,\ldots,x_n\}$. For any $1\leq k\leq n$, choose $A_k\in\mathcal{A}$ containing $x_k$ (such an element of $\mathcal{A}$ exists since $\mathcal{A}$ covers $X$). Then $\{A_0,\ldots,A_n\}$ is a finite subcover of $X$. Thus $X$ is compact.

Note that given any non-empty open set $U$, there are only finitely many points outside of $U$. So we only need a finite number of open sets to get a covering of $\Bbb R$.

The same goes for any non-empty subset of $\Bbb R$. If $\cal U$ is an open cover of some set $A$, then picking any open set $U$ in $\cal U$, the remainder of the entire space (and in particular $A\setminus U$) is finite. So we can find some finitely many open sets from $\cal U$ to cover this remainder.

Therefore every open cover has a finite subcover. So $A$ is compact, but $A$ is arbitrary, so every subset of $\Bbb R$ is compact.