We use engineering strain even though it is not the "correct" value because in most cases, specifically in the elastic regime, engineering strain differs negligibly from true strain.
For linear elastic, Hookean materials, it is generally the case strain at the elastic limit is very small. Even the strongest steels, for example, have an upper limit when cold worked of about $\sigma_{\textrm{el}}=1\times 10^{9}\ \textrm{Pa}$. The modulus of steel is approximately $E=200\times 10^{9}\ \textrm{Pa}$. Thus $\varepsilon_{\textrm{el}}=0.005=0.5\%$ for the strongest steels. So at the onset of plastic deformation, engineering strain is $0.5\%$. Many useful elastic materials have much lower engineering strain at their elastic limits.
For an isotropic, Hookean elastic solid, the following is true
$$
\varepsilon_{x_{1}} = \frac{1}{E}\left[\sigma_{x_{1}}-\nu\left(\sigma_{x_{2}}+\sigma_{x_{3}}\right)\right]
$$
without loss of generality in choice of $x_{i}$. So in uniaxial tension at the elastic limit, $\sigma_{x_{2}}=\sigma_{x_{3}}=0$ assuming the material is free to contract. Thus $\varepsilon_{x_{2}}=\varepsilon_{x_{3}}=-\frac{\sigma_{\textrm{el}}\nu}{E}=-\nu\varepsilon_{\textrm{el}}$. Since the Poisson ratio $\nu$ is approximately 0.3 for steels in the elastic regime, the cross-sectional linear compressive strain is $0.0015$. The cross-sectional area at the elastic limit is thus $(1-0.0015)^{2}A_{0}$, or very close to $0.997$ times the original area.
Thus the true strain is $\frac{1}{0.997}$ times larger than engineering strain at the elastic limit, or about $1.003$ times, or about $0.3\%$ larger. Keep in mind this is at the elastic limit of an exceptionally strong linearly elastic material, and so is a reasonably conservative estimate of the difference between true strain and engineering strain in the elastic regime.
While the above analysis is reasonably useful for linearly elastic, Hookean solids, it doesn't hold quite as well for polymers and biological materials. Such materials are usually viscoelastic (or another class of material entirely), and thus obey different rules in their behavior. True strain also diverges quite wildly from engineering strain in the plastic regime, as evidenced in the following plot (found here)
As for your points:
Measuring changes in cross-sectional area during deformation is difficult. It requires careful placement of calibrated instrumentation on precisely machined test samples. One could use strain gages placed on the sides of a tensile bar to measure lateral strain in uniaxial tension and compression in tensile testing equipment. Obtaining statistically meaningful results takes many samples, as well as significant time, effort and cost.
There is little difference. I hope I have adequately explained above how small the difference is: I calculated approximately $0.3\%$ difference in a conservative case.
The idea that we can ignore anything beyond the end of the elastic regime, or that we always design for the elastic regime, is not true. Plastic deformation can often be worth studying. Modeling continuous shape-forming processes such as rolling, drawing, extruding, etc. requires a deep understanding of the mechanics of plastic deformation to perform successfully, and to that end true stress and true strain are invaluable. Specifically for wire drawing, see (this pdf) and find equation 7. Plastic deformation is also useful for modeling materials that must permanently deform in some expected use cases, such as car body panels and frame components during a collision. The plastic deformation is useful because it absorbs kinetic energy.
Edit: I apologize, I didn't actually answer the question for stress. However, it should be fairly clear that the same points apply to stress as apply to strain given their linear relationship in the elastic regime. Again, in the plastic regime, there can be large variations.
