Impulse response of ideal filters
Given that you are ok with the impulse response of the ideal zero phase lowpass real filter, then you can easily derive the impulse response of the ideal high pass zero phase real filter. In continuous-time let $$h_{lpf}(t) = \frac{\sin(\omega_0 t) }{\pi t} $$ be the impulse reponse of an LTI ideal zero phase lowpass filter whose frequency response is by definition:
$$ H_{lpf}(\omega) = \begin{cases}  1 ,&amp; \text{ for } |\omega| &lt; \omega_0 \\ 0 ,&amp; \text{ otherwise } \\ \end{cases} $$  

Then you can define the frequency response of the ideal zero phase highpass filter as: 
$$ H_{hpf}(\omega) = \begin{cases}  1 ,&amp; \text{ for } |\omega| &gt; \omega_0 \\ 0 ,&amp; \text{ otherwise } \\ \end{cases} = 1 - H_{lpf}(\omega)$$  

By taking the inverse Fourier transform of $H_{hpf}(\omega)$ you can conclude that
$$ \boxed{ h_{hpf}(t) = \mathcal{F}^{-1}\{ 1 - H_{lpf}(\omega) \} = \delta(t) - h_{lpf}(t) }$$

Similarly in discrete-time you can conclude that:
$$ \boxed{ h_{hpf}[n] = \mathcal{F}^{-1}\{ 1 - H_{lpf}(e^{j\omega}) \} = \delta[n] - h_{lpf}[n] }$$