I want to show that $O(n)$ acts freely and properly on $U(n)$ and am trying to use the following.
If $H$ acts freely and properly on $G$, then $H$ also acts freely and properly on any subgroup of $G$.
My argument seems to show that this is true, but this feels like too strong of a statement to be true.
If $H\times X\rightarrow X$ is free, then $hx=x$ implies that $h$ is the idenity element. So clearly for any subset $X'$ of $X$ the action $H\times X'\rightarrow X'$ is also free.
Also if $f:X\rightarrow Y$ is a proper map, and $B$ any subset of $Y$, then the restriction $f:f^{-1}(B)\rightarrow B$ is also a proper map.