Assume that we are in the following situation: a compact Lie group $G$ acts on a compact space $X$ which is not necessarily Hausdorff. $X$ is assumed to be compactly generated and weakly Hausdorff, though. The quotient space $X/G$ is Hausdorff.

Does this imply that $X$ is Hausdorff? I do not think this is true but I'm also unable to find a counterexample since I cannot come up with a non-Hausdorff space that admits a non-trivial $G$-action in the first place.

The reason I'm thinking about this is the following: If the $G$-action on $X$ was additionally free (and $X$ was Hausdorff), then $X$, being compact and Hausdorff, would be completely regular, and hence the quotient map $X \to X/G$ would be a fiber bundle by a result of Bredon. I'm also interested in whether $X \to X/G$ is a fiber bundle if $X,G,X/G$ are as above and the $G$-action is free.

2more comments