There are in fact two field lines that depart each charge headed towards the other. These lines meet at the origin (the mid-point of the two charges), where the field is zero, and vanish there. There are also two other lines, which are born at the origin and depart along the vertical axis. Thus, formally, two lines go in and two lines go out, so no lines actually die in empty space.
These lines are actually a limiting case of lines that leave the point charges at a small angle $\epsilon$ from the intercharge axis; these lines make increasingly close approaches to the origin as $\epsilon\rightarrow0$, and then they shoot off to infinity, increasingly close to the vertical axis.
(If you're sharp, you'll notice there's actually an infinity of such lines, since there's also lines that go off perpendicularly to the screen and at any angle in between. Thus my "two-for-two" argument is not actually quite right. Can you see the limiting behaviour that makes it right?)
Pictures of this were relatively hard to find, but you can see them in this Wolfram web app:
You also have to consider one key point: at the origin, the field is zero, so actually there should be no field lines through it. Or, more formally, the density of field lines should be zero. This comes about in that the angle $\epsilon$ should be really small for the lines to actually approach the origin. You should then plaster the diagram with lines leaving equiangularly at angle $\epsilon$ from each charge, and that will mean a lot of lines on the "outside" of the charges.
Ultimately, though, the lesson is that individual field lines are not that important, and it is the set of lines, equiangularly leaving the charges (in 3D!), that makes a physically relevant diagram. And even then, field line diagrams are only of limited utility in understanding electric fields, mostly because they only incorporate with the utmost difficulty the superposition principle, which is at the real heart of classical electromagnetism.
If you take a permanent magnet, and place a sheet of paper over it. Now sprinkle iron filings on it, and you pretty much get this diagram. This has been the mainstay of field theory since Faraday's time.
A test charge at rest will begin to move in the direction of the field line. Since there is nowhere that it can rest where there is more than one possible direction of motion, there must be no crossings of the field line.
The line that disappears to infinity in one way, and reappears from the other side, means simply that the flux is moving on ever-large circles, and that in the axis-line of the dipole, it is feeding flux as a stream through it. But all this means is that it is turning something that is already there, but never getting a full rotation of the disk up.
In the real world, these polar flux lines simply wander off to another electrical system. Gauss's flux law says that there is a sphere with a net flux across it equal to the enclosed charge: a net of zero does not mean everywhere zero.
Best Answer
Dipole
$\def\vp{{\vec p}}\def\ve{{\vec e}}\def\l{\left}\def\r{\right}\def\vr{{\vec r}}\def\ph{\varphi}\def\eps{\varepsilon}\def\grad{\operatorname{grad}}\def\vE{{\vec E}}$ $\vp:=\ve Ql$ constant $l\rightarrow 0$, $Q\rightarrow\infty$. \begin{align} \ph(\vr,\vr') &= \lim_{l\rightarrow0}\frac{Ql\ve\cdot\ve}{4\pi\eps_0 l}\l(\frac{1}{|\vr-\vr'-\ve\frac l2|}-\frac{1}{|\vr-\vr'+\ve\frac l2|}\r)\\ &=\frac{\vp}{4\pi\eps_0}\cdot\grad_{\vr'} \frac1{|\vr-\vr'|}\\ &=\frac{\vp\cdot(\vr-\vr')}{4\pi\eps_0|\vr-\vr'|^3}\\ \vE(\vr)&=-\grad\ph(\vr)\\ \vE(\vr) &= \frac{1}{4\pi\eps_0}\l(\frac{-\vp}{|\vr-\vr'|^3}+3\frac{\vp\cdot(\vr-\vr')}{|\vr-\vr'|^5}(\vr-\vr')\r) \end{align}
Field Lines
We calculate the field lines with $\vp=\ve_1$ and $\vr'=\vec0$. For calculating the field lines we may multiply the vector field with any scalar field called integrating factor. We choose $4\pi\eps_0|\vr|^3$ as integrating factor and get the differential equation \begin{align} \begin{pmatrix} \dot x\\ \dot y \end{pmatrix} &= \begin{pmatrix} -1+3\frac{x^2}{x^2+y^2}\\ 3\frac{xy}{x^2+y^2} \end{pmatrix} \end{align} for the field lines. In the complex representation $x(t)+i y(t)=r(t)e^{i\phi(t)}$ this reads as \begin{align} \dot x + i\dot y &= -1 + 3\frac{x(x+iy)}{x^2+y^2}\\ &=-1 + 3\cos(\phi)e^{i\phi}\\ \dot r e^{i\phi} + ir e^{i\phi} \dot\phi &= -1 + 3\cos(\phi)e^{i\phi}\\ \dot r + i r\dot\phi &= -e^{-i\phi} + 3\cos(\phi) \end{align} Written in components: \begin{align} \dot r &= -\cos(\phi) + 3\cos(\phi) = 2\cos(\phi)\\ r\dot \phi &= \sin(\phi) \end{align} If we divide these two equations we get \begin{align} \frac{\dot r}{r\dot\phi}=2\cot(\phi) \end{align} We re-parameterize such that $\phi$ becomes the parameter \begin{align} \frac{r'}{r} = 2\cot(\phi) \end{align} Integrate: \begin{align} \ln\l(\frac{r}{r_0}\r) &= \int_{\phi_0}^{\phi} 2\cot(\phi) d\phi = 2\ln\l(\frac{\sin(\phi)}{\sin(\phi_0)}\r)\\ \frac r{r_0} &= \l(\frac{\sin(\phi)}{\sin(\phi_0)}\r)^2 \end{align} As soon as you have some $\phi_0\neq 0$ for $r_0>0$ you get a bounded curve.
Ling-Hsiao Lyu gets the same formula (only that she uses $\phi_0=\frac\pi2$ and $r_0=r_{\rm eq}$) in Section "Dipole Magnetic Field Line" of her Lecture Notes.