Is every regular action proper? Let $A:\,G\times X\to X$ be a continuos action of a topological group $G$ on a topological space $X$.
Question 1: Assume that the action is regular (ie free and transitive). Does that imply that the action is proper? If not: what more do we need for properness?
Question2: Assume that the action is proper and $H\subset G$ is a closed subgroup of $G$. Does that imply that $H$ acts properly on $X$?

For Question 1 I would try to use
$$
A \text{ proper} \iff \psi:G\times X\to X\times X,(g,x)\mapsto (g.x,x) \text{ is proper}.
$$
by showing that $\psi$ is a homemorphism for a regular action. But I dont have a reference for the above equivalence.
For Question 2 I would try to use
$$
A \text{ proper} \iff \{g\in G\vert\, g.K \cap K\neq\emptyset\} \text{ is compact if }K\subset X\text{ is compact} .
$$
Then $\{g\in H\vert\, g.K \cap K\neq\emptyset\}=\{g\in G\vert\, g.K \cap K\neq\emptyset\}\cap H$ is compact. However I dont have a proof for this description of proper actions. Anyone has a reference for that?
 A: As a general note, an action $G\curvearrowright X$ is proper iff for any compact subsets $K,K'\subset X$, the set
$$T(K,K'):=\{g\in G:g.K\cap K'\neq\varnothing\}$$
is compact.  This is only equivalent to the condition at the end of your post with some additional assumptions on $G$ and $X$ (i.e. $G$ is discrete, or $G$ and $X$ are metrizable, etc.).
The answer to your first question is no.  For instance, consider $G=\mathbb R$ with the discrete topology, $X=\mathbb R$ with the usual topology, and the action $\mathbb R\curvearrowright\mathbb R$ given by translation: $g.x=x-g$ for $g,x\in\mathbb R$.  This is free and transitive (these are easily checked), but the action is not proper.  Considering $K=K'=[0,1]\subset\mathbb R$, we have
$$T(K,K)=[0,1]-[0,1]=[-1,1],$$
which is not compact, at least in the discrete topology.
Regarding the second part of your first question, I am not aware of any conditions to add to free and transitive that guarantee the action is proper, other than adding properness itself.
As for your second question, the answer is yes for the reason you state.  As for a reference, I can't think of a good one, but proofs are not too hard to come up with.
