Many of the answers and comments above do not give the whole picture so I believe some clarification is in order. First of all if we take frequency resolution to mean the smallest frequency change where we can find another peak that is distinguishable and not an artifact.
Then the limited resolution is a consequence of limited observation time, capital T in the question.
I'll sketch the proof:
The idea is that what we observe can be described as the real signal $s(t)$ multiplied by a hat window function i.e. $s_o(t)=s(t)[H(t+T/2)-H(t-T/2)]$ where $H()$ is Heavisides step function. The Fourier transform of the hat window is $sin(\frac{Tw}{2})/w/2$, $w = 2 \pi f$. Then the transform is $s_o(w)=s(w)*sin(T/2 w)/w/2$ where * denotes convolution. In other words every single peak in $s(w)$ will trace out the transform of the hat window function centered around that peak and since the transform of the window function is a damped sinus there will be artifacts by the name of side lobes or satellites.
The resolution is commonly taken to be the frequency defined by the first zero of the sinus i.e. $\frac{Tw_{res}}{2} = \pi$ that is $w_{res} = 2 \pi/T$ or $f_{res} = 1/T$. In other words the first point without any contribution from the transform of the hat window is chosen. Note that this is all continuous so far.
Say now that we sample the signal with sampling frequency $f_s$. Then we get $T f_s = N$ samples which then corresponds to N equidistant frequencies in the DFT (discrete Fourier transform) ranging from $-f_s/2$ to $f_s/2$. This means that the step in frequency or frequency bin is $f_s/N = f_s/(T f_s) = 1/T$ which as it happens coincides with the first zero of the sine function above. This is also the smallest non-zero frequency of course. Note that in a power spectrum $1/T$ is the first minimum of the hat window spectrum (not so in the amplitude spectrum).
Now, in practical frequency analysis we don't want side lobes so one usually replaces the hat window with one that does not have any side lobes but this causes degradation of the resolution so in terms of resolution $1/T$ is seldom met in practice but it will always be the first non-zero frequency and frequency bin in the DFT.
Actually, I'm not too keen on the term frequency bin as it might give the impression that the DFT amplitude at a frequency is an integral of the continuous transform over an interval surrounding that frequency but there is no such thing going on. The DFT samples the continuous transform up to $f_s/2$, no more, no less.