If a fiber bundle with a connected total space admits a section, is the fiber connected? Question. If a fiber bundle with a connected total space admits a section, is the fiber connected?

(Since there's a section the bundle is surjective, whence the base is connected as a continuous surjective image of a connected set. Thus the isomorphism type of the fiber is constant over the base.)

For covering maps this reduces to the following.

Fact. For a covering map with connected total space, it admits a section iff it's a homeomorphism.

To prove this one can show the image of any section is clopen upstairs. This approach does not generalize to the case of non-discrete fibers, since sections are generally far from having open image (consider sections of product projections).



As long as we're dealing with a sane base (paracompact Hausdorff), a fiber bundle is a Serre (even a Hurewicz) fibration and therefore has an associated long exact sequence of homotopy groups. This sequence shows that when the total space is connected and the base is simply connected, the fibers must also be connected. Thus an example must involve a non-simply connected base.
I think this follows from the long exact sequence of homotopy groups for a fibration (assuming our bundle is locally trivial). 

Let $F\stackrel{\iota}{\to} E\stackrel{\rho}{\to} X$ be a fibre bundle such that $E$ is path-connected, and let $s\colon X\to E$ be a section, i.e. $\rho\circ s = id_X$. Then for each $n$ we have $\pi_n(\rho)\circ \pi_n(s) = id_{\pi_nX}$, so in particular $\pi_n(\rho)$ must be surjective for all $n$. Therefore the connecting maps $\pi_n X \to \pi_{n-1} F$ in the long exact sequence are all $0$ and so $\pi_n(\iota)$ is injective for all $n$, so in particular $\pi_0F \hookrightarrow \pi_0 E$ so $F$ must be connected.
Not always, a principal bundle has a section if and only if it is trivial.

https://en.wikipedia.org/wiki/Principal_bundle#Trivializations_and_cross_sections