= Profinite topology on the integers is not compact
The <profinite topology> on $\mathbb Z$ has basis the <residue classes> $a+k\mathbb Z$, $k\geq1$. 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 $n\geq1$. 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.
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