From memory the potential barrier for deuterium tritium fusion peaks at around $3$ femtometres.
Suppose we take the deuterium-tritium distance as $r$ then the electrostatic force between the nuclei is:
$$ F = \frac{ke^2}{r^2} $$
and we get can a pressure by dividing this by the area of a sphere with radius $r$ to get:
$$ P = \frac{ke^2}{4\pi r^4} $$
You should regard this as a very rough estimate, but it should be immediately apparent the the $r^{-4}$ dependence is going to be a killer because it rises very rapidly for small $r$. If we take $r$ to be $10$ femtometres we get a pressure of about $10^{28}$ Pa. This is so ridiculously large that even given the rough nature of our estimate it's obvious that this approach is not going to work. The pressure at the centre of the Sun is only around $3\times 10^{16}$ Pa.
2a) I suppose plasma current is mainly carried by electrons which are much more mobile. Then, it is the electrons that are being accelerated by the field. They make (coulomb) collisions with the ions, so ions will gain energy and increase their temperature.
First some background information is necessary based upon some of your comments. A resistivity is a type of drag force that acts to inhibit or limit the relative drift between oppositely charged species, i.e., currents. The drift that leads to currents is a bulk flow, i.e., the first velocity moment of the constituent species (e.g., see discussion at https://physics.stackexchange.com/a/341352/59023 or https://physics.stackexchange.com/a/235549/59023).
The current density in a kinetic gas like a plasma is defined as:
$$
\sum_{s} \ q_{s} \ n_{s} \mathbf{v}_{s} = \mathbf{j} \tag{1}
$$
where $q_{s}$ is the charge of species $s$, $n_{s}$ is the number density of species $s$, and $\mathbf{v}_{s}$ is the bulk flow velocity of species $s$ in the center of momentum frame for the entire plasma (well, technically one can define the current in any reference frame they wish but this would be the physically meaningful one).
Because these electrons are, in the overall process, not gaining any energy, shouldn't their temperature stay the same?
Not necessarily, as I illustrated at https://physics.stackexchange.com/a/268594/59023 it is not required for a plasma to have $T_{e} = T_{i}$ and in many plasmas, there are multiple components per species (e.g., counter-streaming proton beams).
Why is it that $T_{e}$ also increases?
If we assume that the only inhibiting force on the particles is due to Coulomb collisions, then the bulk flow velocities responsible for the currents will be converted to increased thermal speeds to conserve energy. That is, the electrons scatter during the Coulomb collisions as well as the ions, thus it is not just that their bulk flow velocity reduces.
2b) Why is it said that ohmic heating is a loss process when one makes the plasma energy balance?
Because the current shows up on the right-hand side of the electromagnetic energy continuity equation, called Poynting's theorem. The sign of the inner product between the current density and electric field determines whether it is a loss or source mechanism, i.e., if negative(positive) one can say that the particles lose(gain) energy/momentum to(from) the fields.
Because all the energy that is being given to the elctrons in the current by the field stays in the plasma because electrons transfer their energy to the ions and ultimately also to other electrons with whom they can also collide.
Currents act as the source for magnetic fields much like electric charge is a source for electric fields. If the plamsa in which a current exists has a high resistivity, then it will be difficult to maintain currents without significant external work. If the currents are maintained in such a system, then the plasma will heat up (e.g., see https://physics.stackexchange.com/a/348211/59023).
I go into a little more detail about Joule/Ohmic heating at https://physics.stackexchange.com/a/180680/59023.
Best Answer
ITER needs very high ion temperatures (100M K) so the deuterons and tritium nuclei are fast enough to overcome electrostatic repulsion and undergo thermonuclear fusion. A CFL only needs to have a conductive plasma in order to have an electron current exciting atoms in the gas.