Suppose that $f:X\to S$ is a holomorphic morphism of Hausdorff complex manifolds and that $s\in S$ such that $f^{1}(s)$ is compact (and maybe singular). Then is it true that there is an open neighborhood $U$ of $s$ in $S$ such that $f$ is proper over $U$?

No. Let $g:Y\to S$ be an arbitrary finite morphism, $Z\subset Y$ a closed subset such that $Z$ does not contain an entire fibre (for instance any closed point works if the degree of $g$ is larger than $1$). Let $X=Y\setminus Z$ and $f=g_X$. Then for any point $s\in S$, $f^{1}(s)$ is compact (a set of finite points) and nonempty but for any point $s\in g(Z)$ there is no neighbourhood of $s$ in $S$ over which $f$ would be proper. 


Consider $S=\mathbb C^{\star}$, $X=\mathbb C^* \setminus \lbrace 1 \rbrace$ and $f(z)=z^2$. Then $f^{1}(1)=\lbrace 1\rbrace $ is proper, but $f^{1}(U)\to U$ is never proper for any open neighborhood $U$ of $1$. However, if $X, Y$ and $f$ are algebraic, then the answer is yes under the hypothesis $f^{1}(s)$ is nonempty, $f$ is flat (as $X, Y$ are smooth, this comes down to say that all fibers of $f$ are equidmensional of the same dimension), and the fibers of $f$ are connected. See EGA IV.15.7.10. The conclusion is $f^{1}(U)\to U$ is proper over a Zariski open neighborhood $U$ of $s$. There is probably a holomorphic version of this result. EGA III.5.5.2 can also be interesting (over the formal completion of $O_{S,s}$). 

