Why does $\mathrm{sinc}[k]= \delta[k]$ hold?
EDIT: After editing according to the comments, the real question is if and why the following equality holds:

$$\text{sinc}[k]\doteq\frac{\sin(k\pi)}{k\pi}=\delta[k],\quad k\in\mathbb{Z}\tag{a}$$

The answer is, yes, of course it holds. Since $\sin(k\pi)=0$ for any integer $k$, the value of the left-hand side of the above equation must be zero for any integer $k\neq 0$. The case $k=0$ is different because here also the denominator is zero. It can be shown by a limit argument that for $k=0$, the function $\sin(k\pi)/(k\pi)$ equals $1$. Another way to see it is to use the equation that you quoted in your comment:

$$\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{j\omega k}d\omega=\ldots=\text{sinc}[k]$$

If you evaluate this equation separately for $k=0$ you get

$$\frac{1}{2\pi}\int_{-\pi}^{\pi}d\omega=1$$

So we get a sequence which is $1$ for $k=0$ and zero for all other (integer) values of $k$. Consequently, this sequence can be rewritten using the discrete-time unit impulse $\delta[k]$, as shown in Equation (a).

(Below is the original answer explaining that $\delta[k]$ can't be the DTFT of any sequence.)

As mentioned in the comments, this can't be true. Be careful to distinguish between the Discrete-Time Fourier Transform (DTFT) and the Discrete Fourier Transform (DFT). The first is defined by

$$\begin{align}\text{DTFT:}\quad X(e^{j\omega})&amp;=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\\x[n]&amp;=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})e^{j\omega n}d\omega\end{align}\tag{1}$$

The DTFT $X(e^{j\omega})$ of a sequence $x[n]$ is a $2\pi$-periodic function that is continuous in $\omega$.

The (length $N$) DFT is defined by

$$\begin{align}\text{DFT:}\quad X[k]&amp;=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\\x[n]&amp;=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\end{align}\tag{2}$$

Here, $X[k]$ is also a discrete sequence.

The discrete unit impulse $\delta[k]$ can be the DFT of some time domain sequence, but it cannot be the DTFT of any sequence (because it is itself a sequence, not a continuous function). To see which sequence $x[n]$ corresponds to the DFT $X[k]=\delta[k]$, simply use the formula given in $(2)$:

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}\delta[k]e^{j2\pi nk/N}=\frac{1}{N}\tag{3}$$

From $(3)$, it is the constant sequence $x[n]=1/N$ that corresponds to the unit impulse $X[k]=\delta[k]$.