What is Frequency Resolution?
Edit:
I've come to realize that my definition below of &quot;Frequency Resolution&quot; is completely wrong (as well as OP's question). Frequency resolution is how similar the window function's magnitude in frequency space is to the Dirac delta function. This is because the product of the window and the signal in the time domain becomes convolution in the frequency domain (and a convolution with the Dirac delta function is a sampling which would give perfect frequency resolution) The fatter the mainlobe (quantified by it's variance), and the higher the sidelobes, the worse the frequency resolution. Additionally, Time resolution can be quantified a the variance of the window function in the time domain.

Frequency Resolution is not Bin Resolution/Width. In the graph below notice that the lobes dont get closer (frequency resolution) even though bin width is decreasing.

Credit: Dan Boschen
Frequency resolution is rather a property of the Fourier transform of the rectangular function (i.e. the sinc function).

We must window functions to work with Fourier transforms (even when working theoretically). As a consequence we are always working with $f(t)w(t)$ rather than the function $f(t)$ itself (here $w(t)$ is a rectangular function). By the Convolution theorem the Fourier transform of a windowed function is always a convolution of $\hat{f}$ with $\hat{w}=$ sinc. Notably when $f$ is sinusoidal, $\hat{f}$ will be a Dirac delta function and the convolution will just be a sampling of a sinc function. Thus we periodically lose frequencies completely when windowing, the periodicity of this loss is the frequency resolution.
Since, on windowed functions, the DTFT is a periodic approximation of the CTFT it also acquires these properties.
The confusion arises because when we dont pad zeros to the DFT (i.e. only sample $f(t)w(t)$ where $w(t)=1$), the bin width is equal to the Frequency resolution.

However, we may also pad zeros (i.e. also sample $f(t)w(t)$ where $w(t)=0$) and this results in the DTF better interpolating the DTFT of $f(t)w(t)$. Confer with the first graph.

To see why the Fourier transform of the rectangular function is the a sinc funcion watch this video and consider the winding of the sinusoidal functions (its quite involved though)

To answer OP's example the bin resolution is $$\frac{F_s}{N} = \frac{2000}{2000}=1$$ where $F_s=2000$Hz is the sampling rate, and $N$ the DFT size.
The frequency resolution is what the bin resolution would be if we just sampled in the window (no zero padding)
$$\frac{F_s}{M} = \frac{1}{T}=2$$ where $M$ is the number of samples in the window, $T$ is the duration of the sample, and $F_s=M / T$.