In the book of Linear Algebra by Greub, in chapter 8 Section 1, he argues that
(In summary)
$E$ is a real finite dimensional inner product space.
If $\phi: E \to E$, is a normal transformation, i.e for $\bar \phi : E\to E$ $$(x, \phi(y)) = (\bar \phi (x), y)\quad \forall x,y \in E,$$ and $$\bar \phi \circ \phi = \phi \circ \bar \phi.$$ Then $\phi$ and $\bar \phi$ have the same eigenvectors, and any two eigenvector of $\phi$ whose eigenvalues are different are orthogonal.
So from this question we can argue that if $\phi:E \to E$ is a normal transformation and has $n$ eigenvectors whose eigenvalues are different, then those eigenvectors form a orthogonal basis for $E$.
Then in the second section (Self-Adjoint mappings), he first states that
(Direct quote)
It is the aim of this paragraph to show that a selfadjoint transformation of an n-dimensional inner product space $E$ has $n$ eigenvectors which are mutually orthogonal.
And he shows that $e_1$ is an eigenvector of $\phi$.
However, he does not show the existence of such $e_1$, so is it true that either $\phi$ has $n$ eigenvectors which forms a orthogonal basis for $E$, or $\phi$ does not have any eigenvector at all, provided that $\phi$ is self-adjoint. Or the existence of $e_1$ is guaranteed in a way that I can't see ?
I mean, as far as I understood, there can never be the case where $\phi$ has $k$ eigenvectors where $k \leq n$, is it the case ?
Correction:
Reading the next pages carefully, I have realised that after $e_1$, he constructs the remaining eigenvector in the orthogonal complement of $e_1$, and applies the same construction, so if the existence of $e_1$ is not guaranteed, none of the others are also not guaranteed, so $\phi$ can have $k \leq n$ eigenvectors.
Just not to add another question to the unanswered questions queue:
As @Rahul and @Anu pointed out in the comments, if we restrict $F$ to the unit sphere, since unit sphere is a closed and bounded subset of $E$, it is compact, hence $F$ assumes a minimum on that subset.Therefore by definition, $\exists e_1$ satisfying $|e_1| = 1$ s.t
$$F(e_1) \leq F(x) \quad \forall x \quad \text{s.t} \quad |x| = 1.$$
As a result, a self-adjoint mapping of $E$ has to have $n$ eigenvectors (which are mutually orthogonal).