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.

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.