How is every subset of the set of reals with the finite complement topology compact? While we know that every finite set is compact, how to prove that every infinite subset of $\mathbf{R}$, the set of all real numbers, is also compact in the finite-complement topology (also called the co-finite topology)? 
What is the situation in the case of an arbitrary infinite set $X$ with the finite-complement topology? 
 A: 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.
A: 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.
A: A formal proof:
Let $\mathcal{A}\subseteq\tau$ and $\mathbb{R}=\cup\mathcal{A}.$ (We can suppose that $\emptyset\notin \mathcal{A}.$ Why?)
$A\in \mathcal{A}\Rightarrow |\setminus A|<\aleph_0\Rightarrow (\exists x_1,x_2,\ldots,x_n\in \mathbb{R})(\setminus A=\{x_1,x_2,\ldots,x_n\})$
$\left.\begin{array}{rr}\Rightarrow (\exists x_1,x_2,\ldots,x_n\in \mathbb{R})(\mathbb{R}=A\cup (\setminus A)=A\cup\{x_1,x_2,\ldots,x_n\}) \\ \\ \mathbb{R}=\cup\mathcal{A}\end{array}\right\}\Rightarrow$
$\left.\begin{array}{rr}\Rightarrow (\exists B_1,B_2,\ldots, B_n\in\mathcal{A})(x_1\in B_1)(x_2\in B_2)\ldots (x_n\in B_n) \\ \\ \mathcal{A}^*:=\{A,B_1,B_2,\ldots,B_n\}\end{array}\right\}\Rightarrow$
$\Rightarrow (\mathcal{A}^*\subseteq\mathcal{A})(|\mathcal{A}^*|=n+1<\aleph_0)(\mathbb{R}=\cup\mathcal{A}).$
