Find the nyquist sampling rate?
The primary problem in your time domain analysis is the assumption that if both $x(t)$ and $x(t-d)$ have the same Nyquist sampling rate so will $x(t)+x(t-d)$.

You can see that the summation may alter the signal in such a way that resulting signal's bandwidth may be different than the individual ones.

As an example consider the composite signal $x(t) = x_a(t) + x_b(t)$ where $x_a(t)$ is a low-pass signal with individual Nyquist sampling rate of $w_a$ and $x_b(t) = \sin(w_0 t)$, is a high frequency sine wave with a Nyquist sampling rate of $2w_0$. Therefore both $x(t)$ and $x(t-d)$ will have the same Nyquist sampling rate of $2 w_0$.

Then we have $x(t-d) = x_a(t-d) + \sin(w_0(t-d)) = x_a(t-d) + \sin(w_0 t - w_0 d)$. Therefore the sum of those two composite signals, $x(t) + x(t-d)$, will be $$x(t)+x(t-d) = x_a(t) + x_a(t-d) + \sin(w_0 t) + \sin(w_0 t - w_0 d)$$

Now if $d$ is chosen such that $w_0 d = \pi$ then $\sin(w_0 t) + \sin(w_0 t - \pi) = 0$, for all $t$, therefore $x(t) + x(t-d) = x_a(t) + x_a(t-d)$. There will be cancellation of the high frequency terms so that the Nyquist sampling rate of the sum will be $w_a$ (assuming the sum of low-pass signals will not further alter it). Hence the sum will have a different Nyquist sampling rate eventhough $x(t)$ and $x(t-d)$ individually have the same Nyquist sampling rate.