Assuming you want fixed-point free actions...
Since $\pi_1(S^1)$ is cyclic, all connected coverings of $S^1$ have cyclic group of covering transformations. Now, if $\mathbb Z_2^2$ acted without fixed points on $S^1$, the corresponding quotient map would be a covering $S^1\to S^1$ of group $\mathbb Z_2^2$, which is impossible.
Later: a bit less technological: suppose $G=\mathbb Z_2^2$ acts freely on $S^1$ and pick a point $x$. The orbit of $x$ cuts $S^1$ in $4$ segments, and one of them, call it $I$, has $x$ and $\sigma(x)$ as endpoints for some $\sigma\in G\setminus{1}$. But the map $\sigma$ maps $I$ to itself, so by continuity, $\sigma$ fixes a point in $I$, which it didn't.
