Is frequency resolution ("bin size" inverse) always directly proportional to sampling duration (window length)?
For orthogonality of basis vectors, the DFT (or FFT) basis vectors all have a different but exact integer number of full  cycles in the DFT length, e.g. if the FFT result bin for one basis vector has K full cycles, the next bin up will be for K+1 full cycles.

So if you want a DFT to separate 2 frequencies of sinusoids, the closer they are in frequency, the longer a time interval it takes for the two waveforms to become separated by exactly 1 full period or cycle, the longer a DFT (of FFT, or Goertzel, or wavelet, etc.) it takes to fit that time interval for measurement, and have at least a full result bin of separation.

Even if you don’t want to wait until two frequencies of sinusoids differ by one full cycle, say there is so little noise that you can tell apart 2 the sinusoids if they differ by only 0.1 degree (pi/1800.0) of phase change within some window, the rule still applies.  The closer the 2 frequencies are that you want to tell apart or resolve, the longer you will have to wait for them to differ by that 0.1 degree of phase shift or change that you can barely detect (regression fit within some statistical error bound or distribution, etc.)

Otherwise you can’t tell them apart, and thus claim a resolution or level of accuracy that can.

And the more noise, the bigger the difference you will have to wait for (maybe more than 1 degree, maybe even up to more than 2 full periods of difference within the measurement time window).

Note that even with analog filters (or IIRs), generally the narrower the bandwidth, the higher the Q, the longer the impulse response, thus the longer the measurement delay to respond to or detect that narrower frequency range.

So the “tightest” fit requires either more time, and/or low noise (including interference and measurement error), or a very high signal-to-noise ratio (or of the assumption of such, and hope that that assumption is actually correct.)