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Not quite your question, but I'll say it anyway.

The "fixed-points of actions on a $p$-acyclic space are $p$-acyclic" part of Smith theory easily extends to arbitrary $p$-groups. By induction on the order of the group: if $P$ acts on a $p$-acyclic $X$, choose a non-trivial proper normal subgroup $Z \leq P$ (these always exist; if $P$ is non-abelian take its centre), then $X^P = (X^Z)^{P/Z}$.

Let $U(1)$ act continuously on $D^n$, and $X(n)$ be the $\mathbb{Z}/p^n$ fixed points. This is a compact subset, and non-empty by Smith theory. Thus $X=\cap_{n=1}^\infty X(n)$ is also non-empty, by Cantor's intersection theorem. A point $x \in X$ is fixed by the subgroup of $U(1)$ of $p$-torsion p$-power-torsion points; this is dense, so $x$ is fixed by the whole of $U(1)$.
show/hide this revision's text 1

Not quite your question, but I'll say it anyway.

The "fixed-points of actions on a $p$-acyclic space are $p$-acyclic" part of Smith theory easily extends to arbitrary $p$-groups. By induction on the order of the group: if $P$ acts on a $p$-acyclic $X$, choose a non-trivial proper normal subgroup $Z \leq P$ (these always exist; if $P$ is non-abelian take its centre), then $X^P = (X^Z)^{P/Z}$.

Let $U(1)$ act continuously on $D^n$, and $X(n)$ be the $\mathbb{Z}/p^n$ fixed points. This is a compact subset, and non-empty by Smith theory. Thus $X=\cap_{n=1}^\infty X(n)$ is also non-empty, by Cantor's intersection theorem. A point $x \in X$ is fixed by the subgroup of $U(1)$ of $p$-torsion points; this is dense, so $x$ is fixed by the whole of $U(1)$.