It seems to me this is a completely general fact. If $\tilde{\pi}^{-1}(K)$ is contained in a compact open subgroup $K'$, then $\pi (K')$ is a compact open subgroup containing $K$, thus equal to $K$; this implies $K'=\tilde{\pi}^{-1}(K)$.

It seems to me this is a completely general fact. If $\tilde\pi^{-1}(K)$ is contained in a compact open subgroup $K'$, then $\tilde\pi(K')$ is a compact open subgroup containing $K$, thus equal to $K$; this implies $K'=\tilde\pi^{-1}(K)$.