Why closed loop response is faster than open loop

True (for a stable system with normal negative feedback). Negative feedback reduces the time constant of the system, so the output reaches its final value sooner.

Take a first-order plant $G(s) = \dfrac{K}{\tau s + 1}$.

Open loop: for a unit step input

$$c(t) = K\left(1 - e^{-t/\tau}\right), \qquad \text{time constant} = \tau$$

Closed loop (unity feedback):

$$\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{K}{\tau s + 1 + K} = \frac{\dfrac{K}{1+K}}{\dfrac{\tau}{1+K}s + 1}$$ $$c(t) = \frac{K}{1+K}\left(1 - e^{-t(1+K)/\tau}\right), \qquad \tau_{cl} = \frac{\tau}{1+K}$$

The closed-loop time constant is smaller by the factor $(1+K)$, so the response is $(1+K)$ times faster. The price is a lower gain, $K/(1+K)$ instead of $K$, which is made up by adding an amplifier in the forward path.

Example: $\tau = 1$ s, $K = 9$. Open loop: $\tau = 1$ s, settling time $4\tau = 4$ s. Closed loop: $\tau_{cl} = 1/10 = 0.1$ s, settling time $= 0.4$ s, i.e. ten times faster.

For second-order systems the same idea holds: feedback raises the natural frequency $\omega_n$, so rise time and peak time fall (but too much gain can increase overshoot).