Thanks for your comment. I adjusted my question accordingly. To your construction: so whenever $G$ is non-compact this yields an example of a compact PS-manifold $G/\Gamma$ with non-compact isometry group (and if $G$ is compact itself, then the metric on $G$ would be Riemannian), right?

@UriBader: I know that the theorems provided by Kobayashi in his book "Transformation groups in differential geometry" can be applied to Lorentzian mfds as well (e.g. Thm 5.1). So I asked myself why this specific proof of M&S only works for Riemannian mfds.

What you refer to is theorem 1 on the same page. This is a direct conclusion of the embedding into the bundle of orthonormal frames as closed submanifold. Is $(M,g)$ Riemannian and compact then the orthonormal frame bundle is compact since it has the orthogonal group as fibers which is compact. However this is not true in general for Pseudo-Riemannian manifolds. But my question does not refer to this theorem.

Yes this is the paper I had in mind, but I refer to the introduction on page 278, where the author says "We point out that Iso(M,g) has a Lie group structure when considered with the compact-open topology. For Riemannian metrics, this has been established (long ago) in [6]. However, the techniques employed there do not generalize to semi-Riemannian metrics." (The reference [6] is Myers and Steenrod).

@CarloBeenakker: Thanks for the link. I read that question before I wrote mine and I also read the proof of Kobayashi where he embeds the isometry group in the bundle of orthonormal frames as a closed submanifold. I know that this works for Pseudo-Riemannian manifolds too. So I wondered why the original proof from Myers and Steenrod doesn't.