Let's start with the easy one:
A better question would be whether the density of the field lines should be proportional to the field-strength if the field lines start out from the charges at regular angle intervals.
The answer is yes, but: for a 2D plot, when correctly plotted, this is only true for the 2D equivalent of the point charge, i.e. for (combinations of) line charges whose field goes down as $1/r$. This is explained in more depth in the last section of this old answer of mine, but the short of it is that for lines that expand out (or converge in) radially at equal angular intervals, the number of field lines per unit length that cross a given circle of radius $R$ (itself proportional to the field strength) goes down as $1/R$, which corresponds to line charges, not to point charges in 3D.
As for what the field lines should look like, here's a more accurate authoritative plot for the (correctly spaced) field lines of two identical line charges coming out of the page, at a fixed distance but plotted over increasing ranges:
Mathematica source at Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["http://i.stack.imgur.com/2fN3I.png"]
Some salient points in these plots:
- The field lines approach each individual charge at a constant angular separation, as they should.
- When seen at the furthest-out zoom level, the system looks quite a bit like a single line charge of double the charge density, and its field lines reflect that.
- To drill down on that: this requires the field lines to hit the outer circle at an increasingly even separation as the outer circle's radius increases.
- At small zoom levels, however, the field is stronger at the horizontal axis than it is at the vertical ends of the outer circle. (This is because, at the vertical end, the horizontal components of the field cancel out, and the vertical components are not as big.) This means that the line density is larger at the horizontal ends than at the vertical ends.
- In fact, the line density has a local minimum at the vertical ends. This is entirely inconsistent with your first plot, and a surefire way to diagnose it as a buggy plot. (As to what the bug is, it's impossible to tell without seeing the code.)
And finally, an important feature of streamline plots in vacuum in two dimensions: the electrostatic potential there is a harmonic function (i.e. it obeys the Laplace equation $\nabla^2 \varphi=0$), and this means that the streamline plots of its gradient are much easier to find if you see it as a real-valued function of the complex plane, $\varphi:\mathbb C\to \mathbb R$ and find a harmonic conjugate $\chi$ for it, which then acts as its stream function.
Then, because of the nice properties of analytical functions, you know that the streamlines of $\nabla\varphi$ are orthogonal to the gradient $\nabla\chi$, i.e. that $\chi$ is constant over the streamlines. This means that (i) the streamlines are much easier to plot, by simply doing a contour plot of $\chi$, and (ii) that the correct separation in the streamlines can be strictly enforced by simply plotting those contour lines of $\chi$ at regular intervals.
This is the technique used to produce the plots above: here you know that the potential essentially has the form
$$
\varphi(x,y) = \ln\left(\sqrt{(x-1)^2+y^2}\right)+\ln\left(\sqrt{(x+1)^2+y^2}\right),
$$
but that's much easier to express as
$$
\varphi(x,y) = \mathrm{Re}\mathopen{}\bigg[ \ln(x-1+iy) + \ln(x+1+iy)\bigg],
$$
which then immediately gives you the stream function as
$$
\chi(x,y) = \mathrm{Im}\mathopen{}\bigg[ \ln(x-1+iy) + \ln(x+1+iy)\bigg].
$$
Because of the complex logarithms you might have to wrangle with branch cuts and whatnot, but even then, this is way easier (if and when it is available) than the grind of numerically solving ODEs to get the streamlines, where in the general case it is not easy to enforce the correct streamline spacing.
Is there a rigorous mathematical proof of the reason that . . . . .
I cannot give you a rigorous mathematical proof but can give you an indication of what the electric field looks like far away from the charges.
The electric field would be practically indistinguishable from the the electric field due to a single charge $+q$.
The electric field lines would be radial and pointing outwards from the vicinity of the two charges.
The electric field diagram is misleading it does not show that there is a neutral (zero field) point $(1+\sqrt 2)d$ to the right of the $-q$ charge. $d$ is the separation of the charges.
Beyond the neutral point the electric field line is pointing away from the charges (to the right).
To find where the neutral point, $N$, is consider the following diagram.
At the neutral point the electric fields due to the two charges are equal in magnitude and opposite in direction.
$\vec E_{\rm +2q} + \vec E_{\rm -q} = \vec 0 \Rightarrow k\dfrac {+2q}{(d+x)^2} \hat i + k\dfrac {-q}{x^2} \hat i = \vec 0 \Rightarrow \dfrac {2}{(d+x)^2} = \dfrac {1}{x^2} \Rightarrow x = (1+\sqrt 2)d$
I have produced a better set of diagrams to illustrates the points that I have made in my answer. Note the charge of scale from diagram to diagram.
Best Answer
The electric field produced by charge does not consist of discrete field lines. The actual electric force field permeates all of the space around the charge.
For example, the figure below might just as well be used to visualize the relative strength of the field about a point charge (your book fig A), where the degree of shading is intended to convey the field strength, with no gaps. However, shading alone wouldn't provide information on the direction of the field, particularly for more complex charge distributions than a singe point charge, as do discrete field lines with arrows.
The "price" to pay for using discrete lines is the space between the lines that gives the impression (to many others and not just you) that there is no field in the space between the lines.
Hope this helps.