Suppose $⟨·, ·⟩_1$ and $⟨·, ·⟩_2$ are inner products on $V$ such that $⟨v, w⟩_1 = 0$ if and only if $⟨v, w⟩_2 = 0$. Prove that there is a positive number $c$ such that $⟨v, w⟩_1 = c⟨v, w⟩_2$ for every $v, w ∈ V$.
It is the Exercise6.B.11 from Linear Algebra Done Right. I've proved the finite-dimensional case. The "existence of orthonormal basis" theorem can only apply to finite-dimensional inner product space. So how to prove it when it is infinite-dimensional?
Here's a proof that works over $\mathbb{R}$. I am not too used to inner products over $\mathbb{C}$ (as opposed to Hermitian inner products over $\mathbb{C}$), so I am not sure how much of this applies over $\mathbb{C}$.
Pick a non-zero $w\in V$ and define $c:= \frac{\langle w,w\rangle_1}{\langle w,w\rangle_2} = \frac{|w|_1^2}{|w|_2^2}$. Note that $c$ is positive as it is a ratio of positive real numbers.
Lemma 1: If $u\in V$ with $|u|_1 = |w|_1$, then $\langle u,u\rangle_1 = c\langle u,u\rangle_2$.
Proof: Consider the two vectors $u+w$ and $u-w$. Then $$\langle u+v, u-v\rangle_1 = \langle u,u\rangle_1 - \langle u,w\rangle_1 + \langle w,u\rangle_1 - \langle w,w\rangle_1 = 0.$$
So, $\langle u+w, u-w\rangle_1 = 0$. Thus, $\langle u+w, u-w\rangle_2 = 0$. Expanding this gives $|u|_2^2 - |w|_2^2 = 0$, so $|u|_2^2 = |w|_2^2 = c |w|_1^2 = c|u|_1^2$ $\square$
Lemma 2: If $u\in V$, then $\langle u,u\rangle_1 = c\langle u,u\rangle_2$.
Proof: If $u = 0$, the the result is obviously true. If $u\neq 0$, then the element $\frac{|w|_1}{|u|_1} u$ has the same length (with respect to $\langle \cdot, \cdot\rangle_1$) as $w$. By Lemma 1, $\left\langle\frac{|w|_1}{|u|_1} u,\frac{|w|_1}{|u|_1} u \right\rangle_1 = c\left\langle\frac{|w|_1}{|u|_1} u,\frac{|w|_1}{|u|_1} u \right\rangle_2$. Pulling out the factors of $\frac{|w|_1}{|u|_1}$ and canceling gives the result for $u$. $\square$
Lemma 3: If $u,v\in V$, then $\langle u,v\rangle_1 = c\langle u,v\rangle_2$.
Proof: From Lemma 2, $\langle u+v, u+v\rangle_1 = c\langle u +v, u+v\rangle_2$. Expanding both sides gives and using Lemma 2 to cancel $|u|_1^2$ with $c|u|_2^2$ and similarly for $|v|_1^2$ and $c|v|_2^2$, we get $2\langle u,v\rangle_1 = 2c\langle u,v \rangle_2$.