Are eigenfunctions always normed and orthogonal? I came across this simple proof:

We show that Hermitian operators have real eigenvalues. The definition
  of a Hermitian operator is
\begin{equation} \langle \phi_i | \hat A | \phi \rangle = \langle
 \phi_i | \hat A | \phi \rangle^* \tag{1} \end{equation}
Then if $|\psi\rangle$ is an eigenvector of $\hat A$, we have
$$ \hat A|\psi\rangle = \lambda|\psi\rangle \tag{2}$$
and therefore 
$$ \langle\psi|\hat A|\psi\rangle = \lambda . \tag{3}$$
If $\hat A$ is hermitian, we my apply (1) so that
$$\langle\psi|\hat A|\psi\rangle = \langle\psi|\hat A|\psi\rangle^*$$
  $$\lambda = \lambda^*.$$

What I am not getting, is the step from (2) to (3). Seems to me that would be true only if $\psi$ is normed ($\langle\psi|\psi\rangle = \int \psi^*\psi \text d \tau = 1$).
Is it true in general that eigenvectors/eigenfunctions of operators are normed and orthogonal?
 A: The property of orthogonality can always be imposed, but it is not required at all in the excerpt you've cited.
The normalization of the eigenvectors can always be assured (independently of whether the operator is hermitian or not), by virtue of the fact that if $Av=\lambda v$, then any multiple $w=\alpha v$ of that vector will obey
$$
Aw = A\alpha v = \alpha A v = \alpha \lambda v = \lambda w.
$$
Thus, given any eigenvector of any operator, you can always assume (for free) that it's been normalized to unity.
However, this is also not necessary for the manipulations you've cited: if you remove that normalization, then your equation $(3)$ becomes 
$$ \langle\psi|\hat A|\psi\rangle = \lambda \langle\psi|\psi\rangle, \tag{3'}$$
in which $\lambda \langle\psi|\psi\rangle $ is (by the properties of the inner product) a real and positive number. The rest of the manipulations are unaffected: you get to
$$
\lambda \langle\psi|\psi\rangle  = \lambda^* \langle\psi|\psi\rangle
$$
and all you need to do is divide by $\langle\psi|\psi\rangle $.
A: One can always work in an orthonormal basis. By default, we do this when working with quantum mechanics for convenience.
Note that if $\hat{A}|\psi\rangle=\lambda|\psi\rangle,\,\hat{A}^\dagger|\phi\rangle=\mu|\phi\rangle$ then $\langle \phi|\hat{A}|\psi\rangle$ is equal to both $\lambda\langle\phi|\psi\rangle$ and $\mu^\ast\langle\phi|\psi\rangle$, so either the vectors are orthogonal or $\lambda=\mu^\ast$. We can even ensure orthogonality in this special case with a basis change called the Gram-Schmidt process. Finally, we can rescale eigenvectors to have unit norm. This allows such convenient results as $\operatorname{id}=\sum_i |i\rangle\langle i|$ so that $|\Psi\rangle=\sum_i \langle i|\Psi\rangle|i\rangle$.
