What is a proper stack? I have seen the use of the word "proper Deligne-Mumford stack". Now, it is clear to me what it means for a morphism f of stacks to be proper: as usual it should be representable, and every morphism between schemes obtained from f by base change should be proper.
Now, first I guess that "proper" here actually means "complete". A scheme over a field is complete when the structural morphism to a point is proper. But it does not make sense for a stack to ask that the morphism to the point is proper. Indeed it would be in particular representable, and since a point is a scheme this would imply that the stack itself is a scheme.
Another possibility is that the sentence means "a stack with a proper atlas", so that one cannot speak of proper stacks, but only of proper Deligne-Mumford stacks.
So I am asking here what the standard terminology is.
 A: As requested, an answer on terminology
My favorite reference on basics for DM stacks is Edidin's paper, which I find much easier to read than Laumon & Moret-Bailly (who of course deal with Artin stacks).
Short summary. Suppose $P$ is a property of morphisms $f:X\to Y$ in the category of schemes:


*

*If $P$ is local on both $X$ and $Y$ (`local' in appropriate topology, e.g., etale for DM stacks), it makes sense for morphisms of stacks (pass to compatible presentation of both stacks).

*If $P$ is local on $Y$ only, it is easy to define for representable morphisms $F:{\mathcal X}\to{\mathcal Y}$ by changing base to a presentation of ${\mathcal Y}$.

*If $P$ is local on $Y$, but you want to make sense of it for all morphisms, you have to make a special definition --- there is no general approach that works for all properties.
This is what happens with definitions of separated/proper morphism of stacks.
So proper morphism of stacks need not be representable.
A: You can define properness for (not necessarily DM) stacks, this is in Olsson's book and in terms of t3suji's answer is a "special definition".
Definition: (Olsson "Algebraic spaces and stacks", p.210) A map of schemes $f:\mathcal{X}\to\mathcal{Y}$ is proper if it is separated, of finite type and universally closed.

*

*$f:\mathcal{X}\to\mathcal{Y}$ is separated if the diagonal $\Delta: \mathcal{X}\to\mathcal{X}\times_\mathcal{Y}\mathcal{X}$ is proper (as $\Delta$ is always representable, so you can define proper as in point two of t3suji's answer: it means that the pullback of $\Delta$ along $Z\to \mathcal{X}\times_\mathcal{Y}\mathcal{X}$ (for $Z$ a scheme) is a proper map).

*A map $f:\mathcal{X}\to Y$ to a scheme is closed if the image of every closed substack $\mathcal{Z}\subseteq\mathcal{X}$ is closed.

*A map $f:\mathcal{X}\to\mathcal{Y}$ is universally closed if its pullback by any map $Y\to \mathcal{Y}$ (for $Y$ a scheme) is closed.


So for instance, is $BG$ proper? The diagonal map is $BG\to BG\times_{\text{pt}}BG = B(G^2)$, and taking a pullback
\begin{array}{ccc}
G&\xrightarrow{}& \text{pt}\\
\downarrow&&\downarrow\\
BG& \xrightarrow{} & BG\times BG
\end{array}
we see that $BG$ is not proper unless $G$ is proper. I think conversely that $BG$ should probably be proper if $G$ is. More generally, this implies that if $\mathcal{X}$ is a proper Artin stack, the stabiliser groups of points are proper groups.
