Is a closed morphism with proper fibres 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 $f : X\to Y$ 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. 


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

