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.

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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.