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 $f : X\to Y$ 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.

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To answer the question posed as a comment to this answer
(is there a counterexample with connected fibres?? – S.D.):

Let $\alpha:\tilde Y\to Y$ be the normalization of a curve with ordinary singularities, say $Y$ is a nodal cubic curve. For each singular point $P\in Y$, let $Z_P$ denote all but one point of $\tilde Y$ that map to $P$ (choose one randomly that maps to $P$ to exclude). Now let $X=\tilde Y\setminus \left( \cup_{P\in\mathrm{Sing} Y} Z_P \right)$. Then $\alpha: X\to Y$ is one-to-one, closed and have proper and connected fibers, but it is not proper.