Yes, this is correct. To derive it, note that $u$ is $0$ and increasing when $\omega t+\phi_u=0$ and similarly for $y$. The time between these crossing is just as you have said.
No this is not true.
Take the system
$$
\dot{x} = u
$$
which is controllable because its controllability matrix is $1$ and so has full rank. Now take $u=-x$ and you get $\dot{x} = -x$ which is asymptotically stable.
But this doesn't mean we reach the equilibrium in finite time. The solution of the system is:
$$
x(t) = x(0)e^{-t}
$$
So if $x(0) \neq 0$ and you try to solve for $t$:
$$
0 = x(0)e^{-t} \implies 0 = e^{-t}
$$
However the exponential function is never zero for any finite $t$, only in the limit when $t \rightarrow \infty$. So although the system is controllable and the picked controller made the system asymptotically stable, it still takes an infinite time to actually reach the equilibrium, so it is not finite time stable.
Controllability only says that you can reach a state in finite time (for a single time step, with some controller), not that your solution can actually stay there.
For example if you assume $x(0) > 0$ to actually reach $0$ you could just use $u = -1$. Then the solution becomes
$$
x(t) = x(0) - t
$$
So with this controller you reach $0$ after the finite time $t = x(0)$. But you won't stay there.
Best Answer
The delay margin is a measure of robustness (how much delay can be added to the open-loop before it becomes unstable). The closed-loop system should already be stable in order to talk about margins. Why the delay margin can be useful is because a lot of controller applications use a digital implementation, which means AD-converters for the sensors and DA-converters for the actuators. This and the computation of the control logic itself takes time. If were to use a cheaper PLC then this time can increase.
Also it can be noted that the delay margin and phase margin of a closed-loop system are proportional to each other through the bandwidth.