What I understand: By dimension formula $dim V$ $-$ $1$= $dim \langle v \rangle ^ \perp$ since $V$ is the direct sum of $U$ and $U ^ \perp$ if $U$ is a subspace of $V$. dim $\langle v \rangle$ is clearly $1$
So it makes sense that the number of basis vectors for $V$ is $1$ more than the number of basis vectors for $\langle v \rangle ^ \perp$
It also makes sense that if $v$ is an eigenvector then then $e_1$=$v/\|v\|$ is also an eigenvector and also that $\|e_1\|=1$ so $\langle e_1,e_1 \rangle$=1 and $\langle e_1,e_i \rangle$=0 for $i\geq 2$ since $e_i$ were in $v^\perp$ and $e_1$ is a scalar multiple of $v$. So IF $\{e_2,e_3,..., e_n\}$ is a basis of then $v^\perp$ then $\{e_1,e_2,e_3,..., e_n\}$ is a basis for V since an orthonormal set of vectors is also linearly independent.
But WHY can we assume there exists a set $\{e_2,e_3,..., e_n\}$ which is a basis of $v^\perp$. If I can understand this the whole proof will fall together but I am seriously unstuck at the whole line which starts: "So by induction on dim($V$)..." Where is the induction even? Help will be much appreciated.
The inductive statement $P(n)$ is
"For every self-adjoint transformation $T : V \to V$ on a vector space of dimension $n$, there is an orthonormal basis of $V$ consisting of eigenvectors of $T$."
This is clearly true for $n = 0$. To prove it's true in general, we assume $P(n-1)$ for an arbitrary $n > 0$, and show that $P(n)$ follows from this. (That's what the proof you just read used when it said "by induction (i.e,. by assuming that $P(n-1)$ is true) we can construct a basis $v_2, \ldots, v_n$ for $T|_{\langle v \rangle^\perp}\ldots$")
Then the axiom of induction says that $P(n)$ holds for every natural number $n$.