= Solution
Let $A=A^\dagger$ be a <Hermitian matrix> and $Av=\lambda v$ with $v\ne0$. The scalar $v^\dagger Av$ equals its own <complex conjugate>, because
$$
(v^\dagger Av)^*=v^\dagger A^\dagger v=v^\dagger Av.
$$
Since $v^\dagger Av=\lambda v^\dagger v$ and the squared <norm> $v^\dagger v$ is positive, $\boxed{\lambda\in\mathbb R}$.
Now take $Au=\lambda u$ and $Av=\mu v$. The <inner product> satisfies
$$
u^\dagger Av=\mu u^\dagger v,\qquad u^\dagger Av=(Au)^\dagger v=\overline\lambda\,u^\dagger v=\lambda u^\dagger v.
$$
Subtracting gives $(\mu-\lambda)u^\dagger v=0$. Thus distinct <eigenvalues> imply $\boxed{u^\dagger v=0}$, proving <orthogonality> of the corresponding <eigenvectors>. Within a degenerate <eigenspace>, an orthonormal basis may be chosen, but arbitrary distinct vectors in that same <eigenspace> need not be orthogonal.
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