Closed maps with non-compact fibers A geometric example of a non-closed map with compact fibers is obtained by taking a finite covering map and removing a point upstairs. This fragmentation of the fiber destroys closedness but retains compactness of fibers.
What are some geometric examples of a closed map with non-compact fibers? I'm not interested in things like the terminal map of an uncountable discrete space. Hoping for some pictures, the closer to manifolds the better.
My motivation is to justify the intuition for proper (universally closed separated) maps as "complete" families of compact spaces. I want to understand how a "mere" closed map might fail to warrant this description when the fibers aren't compact.
 A: This is not a final answer, but it shows that there are serious restrictions for possible examples.
A map $p : Y \to X$ is not a closed map provided
(1) $X$ contains a countable non-closed set $A$ such that for each $x \in A$ the  fiber $F_x = p^{-1}(x)$ is a closed non-compact subset of $Y$. Note that if all one-point subsets of $A$ are closed in $X$, then the $F_x$ are automatically closed.
(2) $Y$ can be written as $Y = \bigcup_{m=1}^\infty C_m$ with relatively compact open $C_m \subset Y$ such that $C_m  \subset C_{m+1}$. Note that we do not assume that compact includes Hausdorff; if you want to assume this, then $Y$ is nothing else than a Hausdorff locally compact and $\sigma$-compact space. Example: $Y$ locally compact, separable and metrizable.
Proof. Write $A = \{ x_n \}$. For each $n$ the fiber $F_{x_n}$ is closed and non-compact, hence it cannot be contained in any $\overline{C}_m$. Define $Y_n = F_{x_n} \setminus C_n$ (which is nonempty and closed in $Y$) and $B = \bigcup_{n=1}^\infty Y_n$. We show that $B$ is closed in $Y$. In fact, for $y \notin B$ there exists $m$ such that $y \in C_m$. For $n \ge m$ we have $Y_n \cap C_m \subset Y_n \cap C_n = \emptyset$. Hence $D = B \cap C_m = \left(\bigcup_{n=1}^{m-1} Y_n \right) \cap C_m$ is closed in $C_m$ and $U = C_m \setminus D = C_m \setminus B$ is an open neigborhood of $y$ not meeting $B$. But by construction $p(B) = A$.
Remark:
(1) implies that the set $N$ of all $x \in X$ with a non-compact fiber $F_x$ has an accumulation point (in fact, each $\xi \in \overline{A} \setminus A$ is an accumulation point). The converse is true if $X$ is $T_1$ and first countable: If $x$ is an accumulation point of $N$, then there exists a countable $A \subset N \setminus \{ x \}$ such that $x \in \overline{A}$, i.e. $A$ is not closed.
The map $p : (\mathbb Z \times \mathbb R) \cup (\mathbb R \times \{ 0 \}) \to \mathbb R, p(x,y) = x,$ is an example of closed map with infinitely many non-compact fibers.
