A problem with a proof of Bessel's inequality, and how to get Parseval's identity from it I am studying functional analysis, and I think there is a problem
with the proof written in my notes for Bessel's inequality.
The theorem is:

Let $H$ be a Hilbert space and $\{u_{\alpha}\}_{\alpha\in A}$be an
  orthonormal set, then  $$ \sum_{\alpha\in A}|\langle
 x,u_{\alpha}\rangle|^{2}\leq||x||^{2} $$

The proof (in short):

Let $F\subseteq A$ be a finite subset, denote  $$
 V_{F}=SP\{u_{\alpha}:\alpha\in F\} $$
We have see that  $$ \forall x\in H:\, P_{V_{F}}(x)=\sum_{\alpha\in
 F}\langle x,u_{\alpha}\rangle u_{\alpha} $$
[Where we denoted $P_{W}$ as the orthogonal projection on $W$].
Since for every orthogonal projection $P_{W}$ we have that
  $||P_{W}||=1$ [proof included] we have that  $$ \sum_{\alpha\in
 A}|\langle x,u_{\alpha}\rangle|^{2}=||P_{V_{F}}(x)||\leq||x||^{2} $$

I disagree with the last line.
First, I think that it is not true that 
$$
\sum_{\alpha\in A}|\langle x,u_{\alpha}\rangle|^{2}=||P_{V_{F}}(x)||
$$
or that 
$$
||P_{V_{F}}(x)||\leq||x||^{2}
$$
but rather that 
$$
\sum_{\alpha\in A}|\langle x,u_{\alpha}\rangle|^{2}\leq||P_{V_{F}}(x)||^{2}
$$
and then I can say 
$$
||P_{V_{F}}(x)||^{2}\leq||x||^{2}
$$
I think that 
$$
\sum_{\alpha\in A}|\langle x,u_{\alpha}\rangle|^{2}\leq||P_{V_{F}}(x)||^{2}
$$
follows since (I think that): 
$$
||P_{V_{F}}(x)||=\sum_{\alpha\in A}|\langle x,u_{\alpha}\rangle|
$$
and when squaring the sum I have the squares of each term, as well
as some other non-negative elements.
Am I correct ?
How do I get Parseval's identity (equality in Bessel's inequality
under some assumptions about the $\{u_{\alpha}\}$ ) from it ? I don't
know what to say about a product of the form 
$$
|\langle x,u_{\alpha}\rangle|\cdot|\langle x,u_{\beta}\rangle|
$$
I would like it (and all other elements that are not the $\langle x,u_{\alpha}\rangle|^{2}$
to be $0$, but I don't see why this would happen).
 A: It should indeed be $\lVert P_{V_F}(x)\rVert^2$, but with an equality,
$$\begin{align}
\lVert P_{V_F}(x)\rVert^2 &= \left\langle \sum_{\alpha \in F} \langle x,u_\alpha\rangle u_\alpha,\, \sum_{\beta \in F} \langle x, u_\beta\rangle u_\beta\right\rangle\\
&= \sum_{\alpha \in F}\langle x, u_\alpha\rangle \left\langle u_\alpha,\, \sum_{\beta \in F} \langle x, u_\beta\rangle u_\beta\right\rangle\\
&= \sum_{\alpha \in F}\sum_{\beta \in F} \langle x, u_\alpha\rangle \overline{\langle x, u_\beta\rangle}\langle u_\alpha, u_\beta\rangle\\
&= \sum_{\alpha \in F}\sum_{\beta \in F} \langle x, u_\alpha\rangle \overline{\langle x, u_\beta\rangle} \delta_{\alpha\beta}\\
&= \sum_{\alpha \in F} \lvert \langle x,u_\alpha\rangle\rvert^2.
\end{align}$$
Then you have $\lVert P_{V_F}(x)\rVert^2 \leqslant \lVert x\rVert^2$ by $\lVert P_{V_F}\rVert \leqslant 1$.
To obtain Parseval's identity, consider that for any $x \in H$, there is a countable subset $C = \{ \alpha_k : k \in \mathbb{N}\} \subset A$ with
$$\lim_{n\to\infty} \sum_{k=0}^n \langle x, u_{\alpha_k}\rangle u_{\alpha_k} = x,$$
(the necessary condition for Parseval's identity is that the span of the $u_\alpha$ is dense, since only a countable number of coefficients can be nonzero, for each $x$ you get a countable subset $C$), then you can write
$$x = \sum_{k=0}^\infty \langle x, u_{\alpha_k}\rangle u_{\alpha_k}$$
and compute $\lVert x\rVert^2$ like $\lVert P_{V_F}(x)\rVert^2$ was computed above.
