 Profinite topology on the integers is not compact

The profinite topology on $\mathbb{Z}$ has basis the residue classes $a+k\mathbb{Z}$, $k\ge 1$. These classes are both open and closed, since their complements are finite unions of other residue classes. Set $F_n=(4^n-1)/3+4^n\mathbb{Z}$ for $n\ge 1$. Then $F_{n+1}\subseteq F_n$ and every $F_n$ is nonempty. An integer $x$ in their intersection would make $3x+1$ divisible by every $4^n$, forcing $3x+1=0$, impossible. Thus the open complements $\mathbb{Z}\setminus F_n$ form an increasing cover with no finite subcover. The topology is Hausdorff because distinct integers are distinct modulo a sufficiently large integer, but is not compact.