How is this was derived? DTFT for symmetric pulse
First, the following trick will be useful: 
$$ e^{j a} - e^{j b} =  e^{j \frac{a+b}{2}} \left( e^{j \frac{a-b}{2}} - e^{j \frac{b-a}{2}} \right) $$

Then simply expand the sum as:
$$ X(e^{j\omega}) = \sum_{n=-N}^{N}e^{-j\omega n} = \frac{ e^{-j\omega (-N)} - e^{-j\omega (N+1)} }{1 - e^{-j\omega}} = \frac{ e^{j\omega N} - e^{-j\omega (N+1)} }{1 - e^{-j\omega}} $$ 

Apply the trick to the numerator and denominator by recognising $a$ and $b$ carrefully: for the numerator $a=\omega N$ and $b=-\omega (N+1)$ whereas for the denominator $a=0$ and $b=-\omega$ which yields:

$$ X(e^{j\omega}) = \frac{ e^{j\omega N} - e^{-j\omega (N+1)} }{1 - e^{-j\omega}} = \frac{ e^{j \frac{\omega N -\omega (N+1) }{2}} \left( e^{j\frac{\omega N +\omega (N+1) }{2}} - e^{-j\frac{\omega N +\omega (N+1) }{2}} \right) }{ e^{-j\omega/2} \left( e^{j\omega/2}  - e^{-j\omega/2} \right) }$$ 

which simplifies to:
$$ X(e^{j\omega}) = \frac{ e^{-j\omega/2}  \left( e^{j\frac{\omega N +\omega (N+1) }{2}} - e^{-j\frac{\omega N +\omega (N+1) }{2}} \right) }{ e^{-j\omega/2} \left( e^{j\omega/2}  - e^{-j\omega/2} \right) } = \frac{  e^{j \omega \frac{2N+1 }{2}} - e^{-j \omega \frac{2N+1}{2}} }{   e^{j\omega/2}  - e^{-j\omega/2}  }  $$ 

which is:
$$ X(e^{j\omega}) = \frac{  \sin( \omega (N+\frac{1}{2}) )}{   \sin( \omega \frac{1}{2}) }  $$