Discrete time aperiodic signal: Concept of frequency?
TL;DR

Your statements are inaccurate. The concept of frequency is $\frac{1}{T}$ where $T$ is the period time. Be careful with the continuous-time and discrete-time definition for a perioud.




  discrete time sinusoids have frequency −π&lt;ω&lt;π or $−\frac{1}{2}&lt;f&lt;\frac{1}{2}$


So, this statement is not accurate. 

A frequency is defined as the number of times a signal repeats itself in a second and is measured in $[Hz]=[\frac{1}{sec}]$. This is, by definition, not limited to the interval you stated and, infect, not defined well for negative values at all. 

An angular frequency, $\omega=2\pi f$ is the rate of change of angular displacement, θ, (during rotation), or the rate of change of the phase of a sinusoidal waveform. It is measured in angle per second or $\frac{radians}{sec}$. As can be seen from the units, this is also not limited to an interval.

When we start to analyze using the Fourier and Laplace transforms, we analyze in the frequency domain. First, the Laplace transform analyzes the signal and if we want to consider frequencies we convert it to continuous-time Fourier transform (CTFT) by using $H^F(\omega)=H^L(s=j\omega)$, which exist iff $0\in ROC$ for the Laplace transform. Otherwise, CTFT is not defined. 

CTFT is a function of $\omega$ and is not limited to an interval. However, let us look at a $T$ periodical signal $x(t)=x(t+T)$ with a CTFT $X^F(\omega)$.
$$x(t)=x(t+T)$$
$$\mathcal{F}\{x(t)\}=\mathcal{F}\{x(t+T)\}$$
$$X^F(\omega)=X^F(\omega)e^{-j\omega T}$$
$$X^F(\omega)-X^F(\omega)e^{-j\omega T}=0$$
$$X^F(\omega)(1-e^{-j\omega T})=0$$

Either $X^F(\omega)=0$, or $(1-e^{-j\omega T})=0$ and sinc $(1-e^{-j\omega T})=0\implies\omega T=2\pi k$ we know that $X^F(\omega\neq\frac{2\pi k}{T})=0$. This means that the CTFT of a periodic signal is discrete.




  Is the concept of frequency to a-periodic signal whether it is continuous or discrete-time is valid?


You have to be careful here. Is the discrete signal $\sin[n]$ a periodic signal? The surprising answer here is no. A descrit-time periodic signal is defined as a signal for which exist $N\in \mathbb{N}$ S.T. $x[n]=x[n+N]\forall n$. Let us consider $0=\sin[0]=\sin[N]\implies N=2\pi k \notin \mathbb{N}$. on the other hand $\sin(t)$ is most definitely periodical. The concept of frequency in both cases is $\frac{1}{T}$ where $T$ is the time of a single period. For the continuous-time, it is the smallest $T$ for which $x(t)=x(t+T)$ and for a periodic discrete signal, the smallest $N$ for which $x[n]=x[n+N]$. The frequency of both is $\frac{1}{T},\frac{1}{N}$ respectively. 



the discrete-time Fourier transform (DTFT) is the analogy of CTFT for sampled\descrete signals. The difference between both is that it is periodical with respect to the sampling rate. The time distance between every two samples is $T_s$ hance the sampling frequency is $F_s=\frac{1}{T_s}$ and $\omega_s=\frac{2\pi}{F_s}=2\pi T_s$. we now take the original Fourier transform and replicate it around $k\omega_s$ where $k\in\mathbb{N}$. After this operation, we normalize (divide) by $\omega_s$. Now, the DTFT is a function of $\theta$ which is limited to $-\pi \leq \theta\leq \pi$ and is periodic around $2\pi$.
All this can be written as: $$X^f(\theta)=\frac{1}{T_s}\sum_{k=\infty}^{\infty}X^F(\frac{\theta-2\pi k}{T_s})$$

Maybe this is what you meant when you wrote the limitation on $\omega$?



I am not sure where did you get the final relation. Maybe a reference will help and I will edit the answer.