Why is path followed by the test charge not same as the electric field lines?
If the test charge corresponds to a massive particle, the path actually traced out differs from the actual direction of the electric field line, due to inertia.
Assume for illustration that the electric field ${\vec E}$ curves in space, i.e. deviates from a straight line. As this test charge particle moves an infinitesimal distance along the direction of ${\vec E}$, inertia of motion dictates that it would tend to move along that direction itself, even as the ${\vec E}$ for the next infinitesimal element changes direction. Thus, for the second infinitesimal path length, the problem becomes a case of initial velocity along one direction, and the instantaneous acceleration ${\vec a} = q{\vec E}/m$ pointing in a slightly different direction. It is easily imaginable that the trajectory for this infinitesimal path length is not exactly along the direction of ${\vec E}$ itself. The same argument can be readily extended to subsequent infinitesimal length elements.
Thus, any massive particle, left all by itself in an electric field ${\vec E}$ that curves in space, will never trace out the direction of electric field ${\vec E}$. (Please note that there is no such fallacy if electric field points in a straight line, and we launch the test particle along this very straight line!)
Thus, while we may physically like to perceive electric field line as the path taken${}^{1}$ by a test charge particle when launched in the field, in general, there are only two cases in which this procedure is valid (in general, even for curved electric fields):
The mass of the test charge particle tends to $0$. : This way, there would be no inertia, and hence, we avoid the logical fallacy.
Instead of leaving the test charge particle free to move all by itself, there is external involvement. : For example, an "experimenter" would leave the test charge free to move ONLY for an infinitesimal length element, and trace out the path that it takes. Thereafter, the experimenter "holds" the test particle in place, bringing it to rest, and hence, getting rid of the initial velocity for its motion through the next infinitesimal length element. Then again, leaving it free, the path gets traced out. The process can be indefinitely extended to trace out the full path, and hence, the complete direction of electric field (line).
Either way, the bottom-line is, the determination of the electric field line is not that straightforward. Even as a thought experiment, it is smeared with logical pitfalls, and mathematical abstractions!
$^1$ Of course, as the answer establishes, this perception is not strictly true unless one reads the fine print and incorporates any of the additional conditions discussed in the answer. So we are better off visualizing an electric field line as "a line whose tangent at every point points along the direction of the force on a (test) charge", as this would be an accurate description − via this statement, we don't claim that the test charge actually moved.
Edit courtesy : Comment by Farcher, "Footnote" courtesy : John Rennie, in chat.
I would take it to mean that, for the Electric field $\overrightarrow{E}$.
$$\overrightarrow{\nabla}\cdot\overrightarrow{E}=0$$
Everywhere within your charge-free region (lets call it M). Intuitively you can think of a point charge as a place where all of the electric field lines begin or end. picture the field as several arrows, they will point toward or away from the charge, this is what the divergence in the above equation measures.
You could also consider (applying Gauss's Law to the above) to be:
$$\int_{\partial M}\overrightarrow{E}\cdot d\overrightarrow{a}=0$$ where $\partial M$ denotes the boundary of the charge free region. Though this is less helpful as it really means the total charge is zero in M (so you could have charges here so long as they sum to zero).
Best Answer
Your statement implies that the Electric field $\boldsymbol E$ will be parallel, during the whole motion, to the instant velocity $\boldsymbol u(t)$, i.e: $$\boldsymbol E = \boldsymbol E_\parallel + \boldsymbol E_\perp= \boldsymbol E_\parallel.$$ If you have a curved trajectory there must be a component of the force which is perpendicular to the velocity in each instant of time, i.e. the centripetal force:
$$\boldsymbol F = \boldsymbol F_\parallel + \boldsymbol F_\perp \neq \boldsymbol F_\parallel$$ and if the only force is the effect of the electric field on the test charge, i.e. $\boldsymbol F = q\boldsymbol E$, we have that:
$$q \boldsymbol E= q(\boldsymbol E_\parallel + \boldsymbol E_\perp )\neq q \boldsymbol E_\parallel$$
which contradict your statement.