FFT frequency resolution
Suppose you have a signal $x[n]$, with $n \in {0, 1, ... N - 1} $. The same-size DFT is defined by:

$$
X[k] = \sum_{n = 0}^{N-1} x[n] e^{-j\frac{\Large{2 \ \pi n k}}{N}}
$$

The frequency resolution is going to be how many Hz each DFT bin represents. This is, as you have noted, given by $\frac{f_s}{N}$.

If on the other hand you had zero-padded your signal, such that $N_{zp}$ is greater than  $N$, then a more apt term of frequency granularity is given by $\frac{f_s}{N_{zp}}$


  Asking this because the spectrum is symmetric for real-valued inputs.


That is irrelevant. The frequency resolutions/granularities are given by the above. 


  o, say I have fs=1000 Hz and N = 1024, where N is the number of FFT
  points. Now, is the frequency resolution 1000 Hz1024=0.9766 Hz or 1000
  Hz0.5∗1024=1.9531 Hz?


If your sampling frequency $f_s = 1000$ Hz, and you are taking an $N=1024$ (same-sized) FFT, then your frequency resolution is $\frac{1000}{1024}$, which is equal to 0.9766 Hz/bin. If your $N_{zp} = 1024$ (FFT length after zero-padding), then your frequency granularity is 0.9766 Hz/bin.