How can we show that $|H(e^{j\omega})|=|H(e^{-j\omega})|$?
Your statement only holds true for a real sequence $h[n]$.
The frequency response of $h[n]$ equals to
$$H(e^{j\omega}) = \sum_{n=-\infty}^{\infty}h[n] e^{-j\omega n}$$
and if $h(n)$ is real, the conjugate symmetry condition holds
$$H(e^{-j\omega}) = \sum_{n=-\infty}^{\infty}h[n] e^{j\omega n}=\big(H(e^{j\omega})\big)^*$$
So we can derive that
$$\big|H(e^{j\omega})\big| = \big|H(e^{-j\omega})\big|$$
and
$$\arg\{H(e^{j\omega})\} = -\arg\{H(e^{-j\omega})\}$$
where $\arg\{\cdot\}$ represents phase response.
We can also derive that the real part of the frequency response is even symmetric and the imaginary part is odd symmetric.
$$\Re\{H(e^{j\omega})\} = \big|H(e^{j\omega})\big| \cos\big(\arg\{H(e^{j\omega})\}\big) = \Re\{H(e^{-j\omega})\}$$
$$\Im\{H(e^{j\omega})\} = \big|H(e^{j\omega})\big| \sin\big(\arg\{H(e^{j\omega})\}\big) = -\Im\{H(e^{-j\omega})\}$$