What is plotted against the DFT of a windowed sampled signal?
As you've noted, for a length-$N$ time-domain sequence $x[n]$, the $\mathrm{DTFT}$ (aka Fourier transform) is given by the equation:

$$
\mathrm{DTFT}\left\{x[n]\right\}=X(e^{j\omega})=X(\omega), \quad 0\leq n\leq N - 1.\tag{1}
$$

And the $N$-point $\mathrm{DFT}$ is given by uniformly sampling equation $(1)$ to obtain $N$ equally spaced frequencies $\omega_k = 2\pi k/N$: 

$$
\mathrm{DFT}\left\{x[n]\right\}=X(e^{j\omega_k})=X\left(\frac{2\pi k}{N}\right)=X[k], \quad 0\leq k\leq N - 1.$$


If you plot $X[k]$, you're plotting the $N$ equally spaced samples from the $\mathrm{DFT}$ equation not against the continuous frequency $\omega$ but against the index number $k$ which in frequency corresponds to $2\pi k/N$. 
The $\mathrm{FFTs}$ are algorithms to compute the $\mathrm{DFT}$. I don't know what's the goal in plotting the $\mathrm{DFT}$ against $k/F_s$. If you check the units you see that $k$ is just a bin number, unit-less, but $F_s$ is in Samples per second. So, that division has units in Seconds per samples... 
The spacing between the discrete frequencies is indeed $2\pi/N$ radians. By the continuous-time frequency spacing, I think he meant the corresponding frequency spacing in $\rm Hz$ and this is the frequency resolution:
$$f_k=k\cdot \frac{F_s}{N}$$
From here you can guess what's the frequency spacing since two consecutive bins from the $\mathrm{DFT}$ samples are at $\Delta k=1$.