Effect of changing sample rate, window duration and zero padding on DTFT and DFT
halving the $F_s$ (keeping $T$ and $N$ constant) would remove the right half of the bins (halving the maximum frequency)


No. Number of bins remain same because you have not changed $N$. What will change is the resolution. Each bin will represent $F_s/(2N)$ instead of $F_s/N$. Your DFT will not represent analog frequencies from $F_s/2$ to $F_s$ That is why you misunderstood as 'remove right half of the bins'.



  halving $T$ would change the shape of the DTFT, halving the number of lobes. Without zero padding this would halve $N$, but $F_{max}$ would remain constant and so every second bin would be removed


I am assuming that when you halve $T$, you are also reducing DFT to $N/2$. If you reduce FFT size with no zero padding, yes, you are dropping every second bin. Because the resolution now increases to $F_s/(N/2)$. For a period signal, with $N/2$ multiple of period, you will still see only 2 peaks for the FFT because the lobe width also has doubled, and the zero crossing appear exactly at multiples of $F_s/(N/2)=2 F_s/N$.



  doubling N by zero padding the DFT would keep the DTFT and Fmax constant but would double the number of bins


Yes, zero padding would increase the resolution of DFT. $F_s/(2N)$ is the resolution as compared to earlier $F_s/N$. But remember by zero padding you are not adding any new information about the signal. It just reveals more of the DFT you already have.