A quick proof that any compact Hausdorff abelian group of finite exponent is profinite I know that Edwin Hewitt and Kenneth A. Ross (1970) show:
Any compact Hausdorff torsion group is profinite.
But I don't have the book, the proof seems long and I need only the case of abelian groups with finite exponent.
It would be appreciated if you give me a quick  proof for the case.
 A: [I assume all topological groups are Hausdorff.]
This follows easily from Pontryagin duality.  Suppose $G$ is a torsion compact abelian group.  By Pontryagin duality, continuous homomorphisms $G\to \mathbb{R}/\mathbb{Z}$ separate points of $G$.  But since $G$ is torsion, any homomorphism $G\to\mathbb{R}/\mathbb{Z}$ lands in the torsion subgroup of $\mathbb{R}/\mathbb{Z}$, which is $\mathbb{Q}/\mathbb{Z}$.  Moreover, the image of any continuous homomorphism is closed since $G$ is compact, and any closed subgroup of $\mathbb{R}/\mathbb{Z}$ which is contained in $\mathbb{Q}/\mathbb{Z}$ is finite.
Thus every continuous homomorphism $G\to\mathbb{R}/\mathbb{Z}$ has finite image.  Thus continuous homomorphisms from $G$ to finite groups separate points of $G$, so $G$ is profinite.
(By the way, a torsion compact abelian group actually automatically has finite exponent; see https://mathoverflow.net/a/460/75, for instance.  If you know $G$ has exponent $n$, the above argument can be shortened since you can immediately see that the image of any homomorphism $G\to\mathbb{R}/\mathbb{Z}$ must be contained in the $n$-torsion subgroup $\frac{1}{n}\mathbb{Z}/\mathbb{Z}$, which is finite.)
