Confused on Pulse Filter Bandwidth and Symbol Rate Relation ***Still need help in 2021 - not fully clear still on Jan 20th ***
I am confused about the relation between Sinc and Rectangle transform pair and how that relates the Bandwidth of Pulses, Bandwidth of Zero-ISI Filter and the Symbol Rate.
Using :

$T_s$ = Symbol Time Period
$R_s$ = Symbol Rate
$Sinc(t/T_s)$

Logic of Bandwidth of Sinc or Raised Cosine roll off = 0

Time-Domain Sinc : Every lobe is width $T_s$ but main lobe is $2T_s$ because its symmetrical on zero
Frequency-Domain Rectangle : Positive frequency Width of $\frac{1}{T_s}$ (But its Symmetrical on zero and so BW include negative frequency)
So...BW $= 2\frac{1}{T_s} =2 R_s$

Logic of Bandwidth Rectangle

Time-Domain Rectangle : Width $T_s$ (symmetrical on zero)
Frequency-Domain Sinc : Every lobe width is $\frac{1}{T_s}$ but main lobe is $\frac{2}{T_s}$ because its symmetrical on zero
So... BW $= \frac{2}{T_s} = 2R_s$

But I have read that's not right, the Zero ISI Filter BW = $R_s/2$ when the roll off factor  is zero. Why has my logic above got this wrong?
***Still need help in 2021 - not fully clear still on Jan 20th ***
Nyquist says that you can send up to twice as many pulses (symbols) per second as the channel bandwidth $B$ with zero ISI, so you need $R_s \leq 2B$. That is all there is to it. The sinc pulse has zero excess bandwidth so the bandwidth of the signal is equal to the symbol rate, $B_s = R_s$.
About sampling, Nyquist says that you need to sample at least twice as fast as the bandwidth of the signal, so you need $F_s \geq 2B_s$, where $F_s$ is the sampling rate.
Edit
To more directly address the points in the question...

Bandwidth of pulses: the FT of a perfect rectangular pulse is a sinc function which has infinite bandwidth.

Bandwidth of zero-ISI filter: the FT of a sinc function is a perfect rectangular pulse. If the zeros in the sinc function are at $kT_s$, where $k \in (-\infty, \infty)$, then the bandwidth is $\frac{1}{T_s}=R_s$. In case the time domain signal is purely real, then the FT will be symmetric meaning that the negative frequencies give you no more information then the positive frequencies. This cuts the bandwidth in half to $\frac{R_s}{2}$.