Why does every single band-limited signal in frequency have an infinite time domain and vice versa (As it's a symmetric relation, inf in one is finite in the other).
I understand how a digital signal with an infinite slope needs infinite amount of sine waves to approximate or achieve, hence it being infinite in the frequency domain. However, what about any terminating curve with finite slope throughout?
Either an intuitive or a mathematical understanding would be helpful
Many texts prove that a signal cannot be both time limited and bandlimited. It is quite a deep result and depends on complex analysis, but the shortest proof I know starts with a bandlimited signal \$f(t)\$. It is straightforward to show from the fourier transform that being bandlimited means that \$f(t)\$ is analytic over the entire complex plane, hence if it vanishes on any interval (e.g. \$f(t) = 0 \mathrm{~for~} t>T\$), then it vanishes everywhere. Hence a bandlimited signal cannot be time-limited. The converse is identical. You will find this stuff in more rigorous texts (e.g. Papoulis).
A more handwaving argument would be that if it is bandlimited then you don't change it by multiplying the spectrum by a rectangular window, hence you don't change it in the time domain by convolving with a \$\mathrm{sinc}\$ function - this tends to spread it outside any time window you assume it is contained in.
Your question also touches on discontinuities and slopes. There are a lot of useful results about the rate of roll off depending on the level of discontinuity. From memory, a function with step discontinuities has a spectrum that falls off at \$\frac{1}{f}\$, a continuous function with step discontinuities in the first derivative goes as \$\frac{1}{f^2}\$ and so on. The smoother the function (more continuous derivatives) the more rapid the fall off of the spectrum, but also the more spread out on time.
Some problems can benefit from some of the serious maths out there. For example, given that a signal must be constrained to a certain bandwidth, what waveform concentrates most of the signal energy into a certain time interval? The solution to this is the prolate spheroidal functions — see Papoulis's book Signal Analysis.
why does a single band-limited signal in frequency have an infinite time domain
The assumption here is that if the time was limited it might have a discontinuity and thus an infinite rise time if the samples are ideal.
But this is not the case in real systems with BW limited, so it is assumed to be "steady-state".
Therefore the normal case when a system is analyzed with time-boundaries and bandwidth limits we ignore any discontinuities at the end of the "steady-state"
This is analogous to a simple low pass filter with a unit step function applied. In theory the step can be infinite or finite rise time and the exponential never reaches the unit voltage, but in practical terms with tolerances, the experiment duration can be stopped at 10 T=10RC.
At this point=10T, the residual error is about 144 PPM and the dV/dt has reduced the risetime to and spectrum BW or the peak risetime t=0.115% so could be captured with high accuracy with ~142x the -3dB BW.
So in theory yes, you cannot have simultaneous limited time and spectral Fourier BW but if you have an error tolerance, you can have both.
Just to add to other answers here, a concise way to precisely mathematically state this property is to say that "the Fourier transform of a compactly supported L^2 function on R^n is holomorphic, so---if nonzero---is never compactly supported". This is a part of the Paley–Wiener theorem in functional analysis, which you can look up for a proof.
