OFDM violates sampling theorem
This may sound trivial but the sampling theorem can only be violated in the process of sampling an analog signal. In your question you would like to sample the signal 
$$
x(t) = \sum_{k=0}^{N-1}\mathrm{e}^{\displaystyle -j 2\pi k\Delta f t}
$$
and you state correclty that, in order to not violate the sampling theorem, $f_\mathrm{s}&gt;2(N-1)\Delta f$ must hold. In contrast, OFDM systems usually work with a sampling frequency $f_\mathrm{s}=N\Delta f$ and you may now ask where this discrepancy originates.

If we look at $x(t)$ we notice that it has a one-sided spectrum with discrete frequencies at $0, \Delta f, \ldots, (N-1)\Delta f$. That's an odd choice because the signal bandwidth is $2(N-1)\Delta f$ but only $N$ subcarriers are used. With other words: this system has potentially $2N$ subcarriers but the subcarriers $-N, -N+1, \ldots, -1$ are unused and the corresponding spectrum is wasted. We should rather define $x(t)$ like this:
$$
x(t) = \sum_{k=-N/2}^{N/2-1}\mathrm{e}^{\displaystyle -j 2\pi k\Delta f t}
$$
The absolute maximum frequency is now $\frac{N}{2}\Delta f$ and accordingly $f_\mathrm{s}&gt;N\Delta f$.

For creating a discrete-time OFDM signal, however, $x(t)$ is quite irrelevant. You cannot violate the sampling theorem when generating a discrete-time signal, because it can be considered as already sampled. Furthermore, due to the cyclic property of the DFT, the limits of the sum in the definition of $x[n]$ are correct. Whether they are $0 \ldots N-1$, $-N/2\ldots N/2-1$ or $N+3\ldots 2N+2$ doesn't matter.

So I wouldn't say that the textbook is wrong. But if OFDM is motivated by sampling an analog signal, then $x(t)$ as defined in your answer is a poor choice.