Electric field lines dilemma I recently learnt the concept that electric field lines do not cut each other this statement was proved with the logic that :
" If electric field lines would intersect then there would be ultimately two directions of tangents for that very point which contradicts the laws of electric field lines."
But I'm wondering if there's only one line of tangent associated with the electric field lines then how it would not have two directions with it. Namely;
parallel and antiparallel directions 
 A: You are right that any "field line" can be traversed two directions at any point, namely parallel and anti-parallel; what they mean is two linearly independent directions.
The "law of electric field lines" which is being discussed here is
$$ \nabla \cdot \textbf{E} = \rho. $$
If there is no charge at some point $x$, then $\rho(x) = 0$. The gradient of a vector field being zero means precisely that there are no singularities in the integral curves of the field (which can also be intuited as "electric field lines do not cut each other"). 
I believe the point that was trying to be made is that the electric field in the absence of charge is an assignment of an arrow to every point in space. The integral curves of that field, the curves you get by sticking together those arrows, are called "electric field lines."
If you had field lines that cut each other, that would be kind of like the assignment of two arrows to that point in space, since each curve needs to be made up of those arrows.
A: Field lines actually do intersect, for example the field lines emerging from a point charge intersect at the location of the point charge, which is a singularity. Otherwise they don't intersect because the electric field has a unique direction. Also, field lines have an arrow of direction giving the direction of force a positive test charge experiences in the field.
A: Although the field lines usually do not cross because as pointed by Bobak Hashemi, the gradient is usually not zero, it can happen that they cross: consider the 2D potential $V(x,y) = (x^2-y^2)/2$. The divergence free electric field is 
$$ \textbf{E}(x,y) = \left( \begin{array}{c}
x \\ -y \end{array}\right).$$
The "convergent" electric field lines on the $y$ axis cross the "divergent" lines on the $x$ axis at the origin, where of course the electric field vanishes. Note that there are no charge at the origin (nor anywhere else, this is an academic example with all charges located at infinity).
