Is the $G$-action on a principal $G$-bundle proper? Let $G$ be a Lie group.
If $G$ acts properly and freely on a manifold $P$, then it is well-known that $P \to P/G$ form a principal $G$-bundle. I would like to know the converse: namely
Question: if $P \to X$ is a principal $G$-bundle, is the $G$-action on $P$ proper?
Here we say that the $G$-action is proper if the map $P \times G \to P \times P$ defined by $(p,g) \to (pg,p)$ is a proper map (i.e. the preimage of an arbitrary compact set is compact).
 A: There's probably a more complete and efficient proof out there, but here's a partial answer. Let's show it for trivial bundles $P \cong N \times G$ for some smooth manifold $N$ with trivial action $\theta:G \times( G \times N) \to G \times N$ given by $\theta(g,(h,n))=(gh,n)$. Then we'll wave our hands with some local trivializations for the general case. 
Trivial bundle case: Suppose that the bundle is trivial, i.e. $P \cong G \times N$ and the action is simply $g\cdot (h,n) =(gh,n)$. To see that the $G$ action is proper, it suffices to show that for any compact set $K \subset G \times N$, the set $$G_K =\{g \in G : gK \cap K \neq \emptyset\}$$ is compact; see Prop 21.5 in Lee for proof the equivalence.  Let $H$ denote the image of $K$ under the first-coordinate projection $\operatorname{pr}: G \times N \to G$. If $g \in G_K$, then we have $g \cdot k = k'$ for some $k,k'\in K$, hence 
$$g\cdot \operatorname{pr}(k)=\operatorname{pr}(g\cdot k )=\operatorname{pr} (k').$$
This implies that $g =  \operatorname{pr}(k')(\operatorname{pr}(k))^{-1}\in H^{-1} H$, hence $G_K$ lies in $ HH^{-1}$. Note that $H$ is compact because $\operatorname{pr}$ is continuous, in turn implying that $H^{-1}$ is compact because inversion is continuous. We also observe that $HH^{-1} $ is the image of the compact set $ H \times H^{-1} $ under the continuous map $G \times G \to G$ defined by $(g,g') \mapsto g g'$, so $ HH^{-1}$ is compact. Now it suffices to show that $G_K$ is closed. To this end, consider a sequence $(g_i)$ in $G_K$ that converges to some $g\in G$. Since $(g_i)\subset G_K$, there must exist $(k_i)$ in $K$ such that $g_i \cdot k_i \in K$. The compactness of $H H^{-1}\times K$ implies that we can find a convergent subsequence $(g_{i_j},k_{i_j})$. Since $\lim g_{i_j}=g$, we must have $\lim (g_{i_j},k_{i_j})=(g,k)$ for some $k \in K$. Since $\theta(g_{i_j},k_{i_j})$ is a sequence in the closed set $K$ converging to $\theta(g,k)$, we must have $\theta(g,k) \in K$. Thus  $g$ lies in $ G_K$, and we deduce that $G_K$ is closed. $\square$
In the general case, we can cover any compact set $K \subset M$ with a finite number of open sets of the form $\pi^{-1}(U)$ (for open sets $U \subset P/G$) admitting local trivializations $\phi: \pi^{-1}(U) \to G \times U$. Tinkering with the above proof should allow you to stitch together $G_K$ as a finite union of compact sets of the form $\{g \in G : gA \cap B \neq \emptyset\}$ with $A$ and $B$ compact.
A: I think @squirrel's approach can probably be made to work. But here's another approach that doesn't require dealing with a finite cover.  
Suppose $K\subseteq P\times P$ is a compact set. Let $\pi\colon P\to P/G$ denote the bundle projection. To show that the preimage of $K$ is compact, we'll show that it's sequentially compact (which is equivalent for manifolds). Suppose $\{(p_i,g_i)\}_{i=1}^\infty$ is any sequence in $P\times G$ such that $(p_ig_i,p_i)\in K$ for all $i$; after passing to a subsequence, we may assume $(p_ig_i,p_i)\to (q,p)\in K$, and thus $p_ig_i\to q,\ $  $p_i\to p$, and  $\pi(p_i)\to \pi(p)\in P/G$. We need to show that $\{(p_i,g_i)\}$ has a convergent subsequence. 
Let $x=\pi(p)\in P/G$, and choose a neighborhood $U$ of $x$ over which there exists a local trivialization $\phi\colon \pi^{-1}(U)\to U\times G$.  After discarding finitely many terms of the sequence, we may assume that $\pi(p_i)\in U$ for all $i$.  Write
\begin{align*}
\phi(p_i) &= (x_i,h_i),\\
\phi(p) &= (x,h),\\
\phi(q) &= (x,k),
\end{align*}
so that $x_i \to x$ and $h_i\to h$.  Since $\phi$ is $G$-equivariant, 
$$
\phi(p_ig_i) = (x_i, h_ig_i),
$$
and therefore $h_ig_i\to k$.  Since group multiplication and inversion are continuous, we conclude that $g_i = h_i^{-1}(h_ig_i) \to h^{-1} k$, and finally
$$
(p_i,g_i) \to (p, h^{-1}k),
$$
as desired.
