An extended comment: this is not true for actions on more general locally compact (Hausdorff metrizable) spaces. For instance, let $\Gamma$ be any countable group admitting infinitely many distinct subgroups of prime order $H_n$; write $H_\infty=\{1\}$. Denote $F=\{x_m:m\le\infty\}$ (the 1-point compactification of a countable discrete set). Let $X$ be the set of pairs $(f,y)$ with $f=x_m\in F$ and $y\in \Gamma/H_m$. It is naturally a $\Gamma$-set, where $g(f,y)=(f,gy)$. Make it a topological space so that $(x_m,y)$ is isolated for $m<\infty$ and a basis of neighborhoods for $(x_\infty,g)$ with $g\in \Gamma$ is given by $(V_n(g))$, where $V_n=\{(x_m,g'H_m):m\ge n, g\in g'H_m\}$. This makes the action of $\Gamma$ continuous. It is easily seen to be compact, it's slightly more subtle that it is Hausdorff: this uses the fact that for every $g\neq 1$ in $\Gamma$, there exists $n_0$ such that $g\notin H_n$ for $n\ge n_0$. The quotient of $X$ by the $\Gamma$-action can naturally be identified to $F$ and hence is Hausdorff compact.

Clearly the stabilizers are finite. But the action is not proper: write $y_n=H_n$ consider the compact (open) subset $K$ consisting of those $(x_n,y_n)$, $n\le\infty$. Then if $s_n$ is a nontrivial element in $H_n$, $s_n$ fixes $(x_n,y_n)$, and in particular $s_nK\cap K$ is not empty, and hence the action is not compact.

At this time I'm not sure how to change this into a manifold example but I'm skeptical about the claimed result. We can start from a connected Lie group $G$, consider $G\times\mathbf{R}/\mathbf{Z}$ with the left and right $G$-actions on each leaf, mod out the leaf $G\times\{t\}$ by the action on the right of some finite subgroup $F_t$ depending smoothly on $t$ for $t\neq 0$ and tending to the trivial subgroup $F_0=\{1\}$ when $t\to 0$ (in the sense that for every compact subset $K$ of $G$ there exists $\epsilon$ such that $|t|\le\epsilon$ implies $F_t\cap K=\{1\}$; we can find such guys in $\mathbf{SL}_2(\mathbf{R})$). The issue is to arrange the resulting space (on which $G$ still acts on the left, with finite stabilizers and quotient space $\mathbf{R}/\mathbf{Z}$ to be non-singular on the slice $G\times\{0\}$ (possibly $\mathbf{R}/\mathbf{Z}$ could be replaced with another closed manifold if it helps). The next issue is to find a discrete subgroup of $G$ whose action on the resulting space remains non-proper and with Hausdorff compact quotient...

cocompactif the quotient space $G \backslash M$ is compact. It isproperif the map $G \times M \to M \times M$, $(g,m) \mapsto (g \cdot m,m)$ is proper; when the action is cocompact, this happens if and only if all isotropy groups are finite." $\endgroup$ – Dave Witte Morris Apr 7 '15 at 16:44