How can we be sure, that the dot-product of the four-vectors $U$ and $A$ in special relativity is zero?

One way to look at it is by multiplying by $m$:

$$ mu^{mu}=p^{mu}=(E, vec p) $$

with

$$p^{mu}p_{mu} = E^2-p^2 = m^2 $$

If you add some momentum (and energy):

$$ p^{mu} +delta p^{mu} = (E+delta E, vec p + deltavec p) $$

Then:

$$ (p^{mu} +delta p^{mu})(p_{mu} +delta p_{mu}) = E^2 + 2Edelta E - (p^2 +2vec p cdot vec p) = m^2 $$

Rearranging:

$$(E^2-p^2) + 2Edelta E - 2vec p cdot vec p = m^2 $$ $$2Edelta E = 2vec p cdot vec p $$

$$ delta E = frac{vec p}E cdot vec p $$

Note that:

$$ frac{vec p}E = frac{gamma mvec v}{gamma m} = vec v $$

so

$$ delta E = delta vec p cdot vec v $$

Over the infinitesimal time interval $deltatau$:

$$ delta vec p = vec F deltatau$$

$$ vec v = frac{vec x}{deltatau}$$

so that:

$$ delta E = vec F cdot vec x $$

When you apply a force to accelerate an object, the work done to change the energy exactly balances the force's change to momentum to keep the object on the mass shell, which is equivalent to keeping $u^{mu}u_{mu}=c^2$.