Q:

Characterization of cocompact group action

Wikipedia claims the following:

In mathematics, an action of a group G on a topological space X is cocompact if the quotient space X/G is a compact space or, equivalently, if there is a compact subset K of X such that 
  the image of K under the action of G covers X.

My question is: Isn't this wrong?  It is evident that the existence of such a subset K ensures cocompactness, but I am doubting the other direction.  How could one possibly choose K?  Taking an arbitrary transversal (or its closure) does not work, and I do not see what else could be a candidate. 
By the way: Wikipedia points to a specific page in the Handbook of Geometric Topology.  This page, however, contains only the definition of a cocompact space, not the claimed equivalence. 

A:

I agree with @Bugs that some extra assumptions are needed, although I do not have a counter-examples either.  
Here is an argument assuming that $X$ is locally compact.  For each $y\in Y=X/G$ choose (arbitrarily) a point $y'\in X$ which projects to $y$ (you'd need axiom of choice here).  Now, for each $y'$ pick an open neighborhood $U_{y'}$ whose closure in $X$ is compact.  Projecting the open sets $U_{y'}$ to $Y$ yields an open covering by the sets $U_y$ (images of $U_{y'}$'s).  By compactness, there is a finite set $\{y_1,...,y_n\}\subset Y$ so that the sets $U_{y_i}$ cover $Y$. Then the union 
$$
\bigcup_{i=1}^n cl(U_{y_i})\subset X
$$
is compact and projects onto $Y$. 

A:

This is a somewhat silly counterexample, but here goes:
Let $G$ be a topological group and let $H$ be a dense subgroup.  Then the quotient space $G/H$ has the trivial topology, in particular it is compact. 
If $G$ is Polish (separable and completely metrizable) but not locally compact, then every compact subset is nowhere dense.  If $H$ is a countable dense subgroup of $G$ then no compact $K \subset G$ can be mapped onto $G/H$ because $G = \bigcup_{h\in H} hK$ would exhibit $G$ as meager, contradicting the Baire category theorem. 
For an explicit example, let $G = c_0$ be the additive group of the Banach space of real sequences converging to zero with the supremum norm and let $H$ be the countable subgroup of rational sequences with only finitely many non-zero terms. 
This also gives a negative answer to a question raised in the comments: metrizable spaces are compactly generated, so assuming that $X$ is compactly generated is not sufficient for Wikipedia's characterization of cocompactness to hold. 

A:

Here is a counterexample in which $X/G$ is also Hausdorff. 
We give $X:=(\mathbb{N} \times \mathbb{Z}) \cup \{\infty\}$ a topology similar to the Arens-Fort topology.  That is, each point $(m,n) \in \mathbb{N} \times \mathbb{Z}$ is isolated and basic neighborhoods of $\infty$ are of the form $$B_{f,k}:=\{\infty\} \cup \{(m,n) : n \geq f(m) \land m \geq k\},$$ where $f:\mathbb{N} \to \mathbb{Z}$ and $k \in \mathbb{N}$. 
We let $G:=\mathbb{Z}$ act on $X$ in the natural way: $g(m,n)=(m,n+g)$ and $g(\infty)=\infty$.
Any compact subset of $X$ is finite, so its orbit cannot cover $X$. On the other hand $X/G$ is just a convergent sequence (which is compact and Hausdorff). 