why can the system reach its final value if there is always a control signal active?
For a system where \$y(t) = 10u(t)\$ it is easy to see that to obtain any nonzero \$y(t)\$ you need a nonzero \$u(t)\$. In the case of a dynamical system, remember that \$Y(s) = G(s)U(s)\$ in the time domain is
$$ y(t) = g(t)y(0)+\int^t_0 g(t-\tau)u(\tau)d\tau,$$
For the system
$$ G(s) = \frac{1}{s+2} \Longrightarrow g(t) = e^{-2t},$$
Even if you start with a nonzero \$u(t)\$, if from an instant \$t_0\$ onward you have \$u(t)=0, \; t\geq t_0\$, that would lead to
$$ y(t+t_0) = g(t)y(t_0) + e^{-2t}\int^t_{t_0} e^{2\tau}u(\tau)d\tau = g(t)y(t_0),$$
in cases where you have a stable system \$G(s)\$ this will mean
$$ y(t+t_0) \xrightarrow{t\rightarrow \infty} 0.$$
So, having \$u(t)=0\$ from a point onward in a stable system will lead to the steady-state of \$y=0\$.
One case where you could have a \$u(t)=0\$ at the steady-state is if your system is an integrator, with
$$G(s)=\frac{1}{s}.$$
Or is this the reason why overshoot happens?
As you have mentioned, as you have the overshoots, and undershoots, you will have that \$e(t)\$ goes from positive to negative and so on. When \$e(t)>0\$, \$u(t)\$ is increasing, and for \$e(t)<0\$, \$u(t)\$ decreases.
So I used again the example system
$$ G(s) = \frac{1}{s+2}$$
and the control
$$u(t) = 10\int^t_0(r(z)-y(z))dz$$
which resulted in the following step response. Notice that at all those red boxes we have \$e(t)=0\$, and they are the inflection point of \$u(t)\$, but none of them are the steady-state (when y(t)=r(t) and stays so for any future time). And that should point that your remark about "[u(t)] is still there and larger than 0, while the P-action and D-action both are zero and have no effect anymore." is only correct at the steady-state, because at most points where \$e(t)=0\$ the P-action will be zero, but not the derivative one.
Is it that the error gets smaller once e(T)=r(T)−y(T)<0 and the integral gets smaller?
First it would be better to say the "integral gets closer to the steady-state control", since it doesn't always means getting smaller.
There will be situations where the error will not decrease after the inflection point, specially if there are delays. But for the system that I used as an example it does so.
