ag.algebraic geometry - morphism closed + fibres proper => proper?

The answer is no. Consider an integral nodal curve $Y$ over an algebraically closed field, normalize the node and remove one of the two points lying over the node. Then you get a morphisme $f : Xto Y$ which is bijective (hence homeomorphic), separated and of finite type, and the fibers are just (even reduced) points. But $f$ is not proper (otherwise it would be finite and birational hence coincides with the normalization map).

In the positive direction, you can look at EGA, IV.15.7.10.

[Add] There is an elementary way to see that $f$ is not proper just using the definition. Let $Y'to Y$ be the normalization of $Y$. So $X$ is $Y'$ minus one closed point $y_0$. It is enough to show that the base change of $f$ to $Xtimes Y' to Y times Y'$ is not closed. Consider the closed subset
$$Delta=leftlbrace (x, x) mid xin X rightrbrace subset Xtimes Y'.$$
Its image by $f_{Y'} : Xtimes Y' to Ytimes Y'$ is $leftlbrace (f(x), x) mid xin Xrightrbrace$ which is the graph of $Y'to Y$ minus one point $(f(y_0), y_0)$. So $f$ is not universally closed, thus not proper.