About divergence-free field lines
It isn't necessary for the lines of a divergence-free field to be closed. Zero divergence implies that the field lines can't end. Closed curves don't have endpoints, but there are other possibilities, such as lines that extend infinitely in both directions.
The relevant mathematical result here is Gauss's theorem. For any field, it says that the integral of its divergence in any region of space equals the integral of the field over the boundary of the region, the flux of field lines crossing the boundary surface. If you enclose in such a surface a point at which field lines start or end, the flux doesn't vanish, so the divergence of the field can't be zero.
For magnetic fields, the divergence is zero and then there aren't any ending points for its lines. For electric fields, the divergence equals the density of charge, and so its lines have ends at point with non-vanishing charge. The flux of lines across a closed surface is proportional to the charge enclosed by it.
A geometrical picture of irrotational fields
So we have a picture of electric field lines being born and dying at some (charged) points or regions of space. But we haven't used the property that it's irrotational.
If you're looking for some geometrical implication of zero rotational, a way of getting it is considering the surfaces that are orthogonal to field lines at every point. The existence of this surfaces, even locally, is not guaranteed if our field is not irrotational. However, they can be defined for some fields with non vanishing rotational. For example, for the magnetic field around an infinite straight wire, they would be the semiplanes limited by it. A surface defined in this way has an orientation: at each point one of its two orthogonal directions is determined by the direction of the field at that point. Although it's not a standard terminology, let me call them here "field surfaces". A nice analogy with field lines in the previous case will arise.
We can imagine field surfaces having boundaries; that is, there are curves in space at which these surface can end. For example, you can picture such a curve and a family of surfaces sharing it as a boundary, just as a family of field lines might end at a point. If we take an oriented closed curve wrapping around a boundary curve, the integral of the field over it will be proportional to the number of field surfaces crossing it, including a negative sign for each that has opposite orientation as the curve an a positive sign otherwise.
There's a generalization of Gauss's theorem that's useful in this case. It's called Stokes' theorem. The version we need now says that the integral over a surface of the rotational of a field equals the integral of the field over the boundary of the surface, a closed curve. Suppose that the surface over which we are integrating cuts a boundary curve. Therefore, by integrating the rotational we are counting the field surfaces that cut its boundary.
For electric fields (in the static case) the rotational is zero. Thus, following the reasoning above, we can conclude that electric field surfaces can't end.
A more precise formulation of the second part
The picture above can be formulated more precisely using differential forms and the Frobenius theorem. The idea is not to use the fact that a divergence-free field has a potential, but to work with geometrical objects which might be understood intuitively, as field lines are. In particular, we want to understand what theses field surfaces are and to learn something about their existence.
In general, any non-vanishing $1$-form $\omega$ on a $n$-dimensional manifold $M$ defines at each point $p\in M$ a $(n-1)$-dimensional subspace $S_p$ of the tangent space $T_pM=\mathbb{R}^3$ as $S_p=\{v\in T_pM:\;\omega(v)=0\}$. Such a smooth assignment of subspaces of the tangent space to every point is called a distribution.
One question we can ask about a distribution is whether we can find a family of $(n-1)$-dimensional submanifolds such that their tangent space at each $p$ is $S_p$. If we can, the distribution is said to be integrable. For our case, we will need a particular case of the Frobenius theorem: the statement that if a differential form is closed, then its associated distribution is integrable.
Now, the electric field can be seen as a $1$-form $E=E_xdx+E_ydy+E_zdz$ in the manifold $M=\mathbb{R}^3$. The distribution associated with $E$ is nothing more that the family of planes given by orthogonality to the electric field at a point.
The condition that its rotational is zero is then just $dE=0$. So by the Frobenius theorem, the ($2$-dimensional) surfaces that are tangent to the planes $E$ defines exist. These are just the "field surfaces" that are used above.
Best Answer
As stated in the other post linked, there are background solutions to maxwells equations, which are determined independantly to charges and currents.
One of which COULD be a constant electric field. When I say constant, I mean the electric field has the same single vector attached to every point in space. ( overplayed onto of fields produced by charges and currents)
IF this homogenous solution was a constant E field, the associated background B field would be a time independant function too. As
$\nabla × \vec{E} = \frac{\partial \vec{B}}{\partial t}$
$0 = \frac{\partial \vec{B}}{\partial t}$
$\vec{C}(x,y,z) = \vec{B}$
This homogenous constant E field solution to maxwells equations would be different then the typical "electrostatic fields" we are used to, because in this solution $\nabla \cdot \vec{E} = 0$. Where as the typical E fields from charges, would have a non zero divergence at some point in space.