I may be misunderstanding the question, but it seems rather straightforward to me.

a) As stated, without assuming, that $X$ is, say, reduced, it is certainly false:
Let $X=\mathrm{Spec}k[\varepsilon]=k[x]/(x^2)$, $Y=\mathrm{Spec} k$ and $f:X\to Y$ the structure map of $X$ as a $Y$-scheme. This is obviously a homeomorphism topologically and it has a section that maps $Y$ isomorphically to $X_{\mathrm{red}}$.

b) It seems to me that pretty much this is the only thing that can go wrong: If $f$ has a section, then by definition of a section, $Y$ is isomorphic to the image of the section.

As $Y$ is regular, it is necessarily reduced, so if $f$ is one-to-one, then $f$ and the section induce an isomorphism between $Y$ and $X_{\mathrm{red}}$. So, if $X$ is reduced, the statement is true, if $X$ is not reduced, then the question is whether there exists a morphism $X\to X_{\mathrm{red}}$ that's an isomorphism on $X_{\mathrm{red}}$. The above example shows that this can happen, so some assumption is required to rule out this possibility.

If $f$ is not one-to-one, something similar would still happen. The section would still induce an isomorphism between $Y$ and the image of the section which would have to be an irreducible component of $X_{\mathrm{red}}$. On the other hand, in this case you would probably want to assume that $X$ and $Y$ are irreducible, since otherwise you can just add an additional component that screws things up. However, then (assuming that $X$ is irreduciblke) this implies that $Y\simeq X_{\mathrm{red}}$ and you end up in the previous case again.

Commutative Ring Theorywhere they prove the result for flat with regular closed fiber. – Karl Schwede Feb 4 '12 at 18:12