$2\pi$ periodicity of discrete-time Fourier transform
The argument does not work in continuous time. In discrete time the argument is that
$$e^{j\omega n}=e^{j(\omega+2\pi)n},\qquad n\in\mathbb{Z}\tag{1}$$
This is true because by definition $n$ is an integer. In continuous time we generally have
$$e^{j\omega t}\neq e^{j(\omega+2\pi)t},\qquad t\in\mathbb{R}\tag{2}$$
because $t$ is a real variable. Eq. $(2)$ only holds with equality for integer $t$, but not for all other values of $t$ (which would be necessary for $e^{j\omega t}$ to be $2\pi$-periodic).
But the fundamental reason why the DTFT is periodic is the fact that we deal with sequences, and if we assume that the sequence is the result of sampling a continuous-time signal, then it becomes obvious that different continuous-time signals can result in the same sampled sequence. E.g., take a continuous-time cosine
$$x(t)=\cos(2\pi f_0 t)\tag{3}$$
Now we sample $x(t)$ at instances $nT=n/f_s$, where $f_s$ is the sampling frequency:
$$x[n]=\cos(2\pi f_0 nT)=\cos(2\pi n f_0/f_s)\tag{4}$$
I leave it as an exercise to you to show that the frequencies $f_0=f_s-\Delta f$ and $f_0=f_s+\Delta f$ (with some $\Delta f&gt;0$) result in the same sequence $x[n]$. This phenomenon is called aliasing. This means that frequencies above $f_s/2$ (also called the Nyquist frequency) cannot be distinguished from the corresponding frequencies below the Nyquist frequency. That's why the DTFT is periodic. There cannot be any new spectral information outside the frequency interval $[-f_s/2,f_s/2]$. If you haven't learned it yet, check out the sampling theorem, which will also help you understand the nature of spectra of discrete-time signals.
Also have a look at this answer to a related question.