Can the circle group $S^1$ act smoothly and freely on the Klein bottle? I'm sure there is some obvious reason why the answer is no, which eludes me right now.

We can view $K$ as the quotient of $S^1\times S^1\subset\mathbb{C}\times\mathbb{C}$ by the involution $(z_1,z_2)\to (-z_1,z_2^{-1})$. Then we get an almost-free $S^1$-action $(z,[z_1,z_2])\to [zz_1,z_2]$ with $\mathbb{Z}_2$ isotropy ($-1$ fixes the circles $[z_1,1]$ and $[z_1,-1]$).