Understanding why (at least) 2 points of the FFT of a real signal are real
First, &quot;why (at least) 2 points of the FFT is real&quot; is incorrect.  It is why one or two points of the FFT is real.
Each point of a discrete Fourier transform (which the FFT is just an algorithm to compute quickly) is the complex coefficient for its corresponding frequency.  In other words, the DFT starts with the assertion that the signal in question can be fully described by
$$x_k = \sum_N a_k e^{\frac{2\pi}{N}k} \tag 1$$
For $x_k \in \mathbb R$, (1) just carries the constraint that $a_k = (a_{N - k})^*$, where $x^*$ is complex conjugation.
From this, we can show that $a_0$ (and $a_{N/2}$ for $N$ even) are constrained to be real:
Start with the rule I mentioned that in order for $x_k$ to be real, $a_k = (a_{N - k})^*$.  Another property of the DFT is that it's only uniquely defined over an interval $N$ points long, and then it repeats.  So $a_{k + N} = a_k$.  This means that for $k=0$, $a_{N - k} = a_{N} = a_0$.  Applying the conjugate property, this means that $a_0 = a_0^*$ -- so $a_0$ must be real.
For $N$ even, $a_{N/2} = a_{N - N/2}$.  So, again, $a_{N/2} = a_{N/2}^*$, so $a_{N/2}$ must be real.
So -- one (or two, in the case of an even number of points) arguments must be real.