**Fact:** For any (continuous) $S^1$-action on the closed unit disk $\mathbb{D}^n$, there is a fixed point $x_0\in\mathbb{D}^n$.

I have thought of a possible argument that re-proves this, but am not sure how to complete it:

Let $U_p\subset S^1$ be the subgroup of $p^\text{th}$-roots of unity ($p$ prime). An $S^1$-action on a compact contractible space $X$ will induce a $U_p$-action on $X$. Smith Theory then implies that $X^{U_p}$ is nonempty, i.e. there is a fixed point $x_p\in X$ under $U_p$, for any given prime $p$. *Now here is where I want to say:* Taking $p$ sufficiently large, we find a fixed point $x_\infty$ under $S^1$. (The intuition is that $\lim_{p\to\infty}U_p\approx S^1$, and denseness will be sufficient by continuity of the action.)

1) **Is it possible to fill this gap, i.e. can this 'proof' make sense?** Not sure how to make sense of this limit/sequence of $U_p$'s, and whether the fixed points hop back and forth forever.

2) **Is such a sequence $\lbrace x_p\rbrace_{p=\text{prime}}$ Cauchy? Or, does there exist a prime $p_0$ where $x_p=x_{p_0}$ for all primes $p>p_0$?**