As mentioned in Wikipedia, a proper morphism between schemes is an analog of a proper map between complex analytic spaces.
Why is it so?
A proper morphism $f:X \to Y$ of schemes is by definition separated, of finite type and universally closed. If we view this morphism as a holomorphic map $f:X(\mathbb{C}) \to Y(\mathbb{C})$, then (I guess) separatedness and being of finite type should imply that each fiber $f^{-1}(y)$ is Hausdorff and finite dimensional, respectively. The "key" feature of a proper morphism is being universally closed, which (I guess) should imply that the pre-image $f^{-1}(A) \subset X(\mathbb{C})$ is compact for any compact subset $A \subset Y(\mathbb{C})$.
If my guesses are correct, then the question becomes: why does universally closedness of the morphism $f:X \to Y$ imply that $f^{-1}(A) \subset X(\mathbb{C})$ is compact for any compact subset $A \subset Y(\mathbb{C})$?