A free irrational torus action that is not proper

Define the $\mathbb{R}$-action on ${\mathbb{T}}^2$ by $t\cdot (z,w)=(e^{2\pi it}z,e^{2\pi i\alpha t}w)\mathrm{.}$ By [F2], it is a smooth free left action.

Every orbit is dense by [F2].

The action is not proper. The map $u\mapsto e^{2\pi iu}$ identifies the wrap-metric circle in [F3] with the complex unit circle $S^1$: their chordal distance is $\mathrm{\mid }{e}^{2\pi iu}-e^{2\pi iv}\mathrm{\mid }=2\sin (\pi d(u,v))$, so the map is a homeomorphism. Thus [F3] makes $S^1$, then ${\mathbb{T}}^2\times {\mathbb{T}}^2$, compact. The full inverse image of this compact target under the action-graph map is $\mathbb{R}\times {\mathbb{T}}^2$. Were it compact, its continuous projection onto $\mathbb{R}$ would make $\mathbb{R}$ compact by [F4], contrary to the open cover $\{(-n,n):n\ge 1\}$, which has no finite subcover.

The quotient is not Hausdorff. Each orbit is a proper dense subset: it is dense by step 2.1. For a point $(z,w)$, choose $\theta \in \mathbb{R}$ with $z=e^{2\pi i\theta }$. Its orbit meets $\{1\}\times S^1$ only at the countable set $\{(1,e^{2\pi i\alpha (n-\theta )}w):n\in \mathbb{Z}\}$, so it cannot contain the whole circle $\{1\}\times S^1$ and is therefore proper. If the quotient were Hausdorff, a singleton orbit class would be closed and its inverse image under the quotient map would be a closed orbit, contradicting density and properness. This free, nonproper action therefore refutes both conclusions, without using any choice principle.