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Here is a simple visualization of the example that Pece is talking about. The top map is a 2-fold covering but the bottom map has infinite fibers. The composition isn’t locally trivial and is therefore not a covering map. However, it is still a semicovering map in the sense that it is a local homeomorphism which has the unique path lifting property.
Yes there is. The following is a classical example (I think it is in Hatcher for example).
Take $H$ to be the hawaiian earings space, that the subspace of $\mathbb C$ which is the union of the circle $C_n\, (n\geq 1)$ of center $\frac 1 {2n}$ and radius $\frac 1 {2n}$.
Then you can get a composition of covering maps which is not a covering map as follow :
I let you fill the details, but the idea is the following : $p \colon E \to H$ should be a map such that for any open neighbourhood $U$ of $0$, $p^{-1}(U)$ is almost a (countable) disjoint union of copy of $U$ except one of the summand screws things up.