Question

Is a closed morphism with proper fibres proper?

Answer 1

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 $f : X\to Y$ 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 $X\times Y' \to Y \times Y'$ is not closed. Consider the closed subset $$\Delta=\left\lbrace (x, x) \mid x\in X \right\rbrace \subset X\times Y'.$$ Its image by $f_{Y'} : X\times Y' \to Y\times Y'$ is $\left\lbrace (f(x), x) \mid x\in X\right\rbrace$ 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.

Answer 2

Any surjective morphism between two curves is closed and have proper fibres. Obviously not all of them are proper.

Answer 3

“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:X→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). ”

I am afraid that f is not closed. We can see it in the following: we choose a neighborhood U of the pre-image (the left point of the two points) of the node, then A=X\U is closed, but the image of A is not closed. Since f(A) doesn't contain the node but contain a branch of Y around the node.