Frequency response of $\mathrm{sinc}[n]$
First of all, there is a mistake in that image: there should be an $n$ in the denominator instead of a $t$.

Check that your signal can be expressed as

$$h[n]=2\frac{\sin\left(\frac{\pi}{2}n\right)}{\pi n}=\frac{\sin\left(\frac{\pi}{2}n\right)}{\frac{\pi}{2}n} \tag{1}$$.

Using Euler's formula we can express that sine as the sum of two exponentials:

$$h[n]=\frac{1}{\pi/2}\frac12\frac{e^{jn\pi/2}-e^{-jn\pi/2}}{jn}=\frac{1}{\pi}\frac{e^{jn\pi/2}-e^{-jn\pi/2}}{jn}$$

Let me multiply and divide that expression by $2$, which doesn't change anything but will be useful later:

$$h[n]=\frac{1}{2\pi}2\frac{e^{jn\pi/2}-e^{-jn\pi/2}}{jn} \tag{2}$$

Notice that there exists an integral that returns that exact function of $n$ and, on top of it, that integral represents the inverse Fourier transform of some function:

$$\frac{1}{2\pi}\int\limits_{-\pi/2}^{\pi/2}2e^{j\omega n} \ \mathrm{d}\omega=\mathcal{F}^{-1}[H(e^{j\omega})]=h[n] \tag{3}$$

where $H(e^{j\omega})$ is periodic in $2\pi$ and, for each period:

$$H(e^{j\omega}) =
\left\{
 \begin{array}{ll}
  2  &amp; \mbox{if } |\omega| \leq \pi/2 \\
  0 &amp;   \ \mathrm{otherwise}
 \end{array}
\right.$$

So the Fourier transform of $h[n]$ is $H(e^{j\omega})$, because the inverse Fourier transform of the latter is the former. And that's the proof.

However, most of the times one does not do all these calculations and just looks up the transform of $\mathrm(1)$. (In the link's table, look up for your signal and replace $W=2$. Then use the linearity of the Fourier transform to get to the result you are looking for.)