Let $(V,\langle\cdot,\cdot\rangle)$ be a real or complex Hilbert space and $\{\psi_1, \psi_2,\ldots\}$ a set of orthonormal vectors from V. Suppose $x \in V$ is such that $\|x\|^2=\sum_{n=1}^{\infty} |\langle x,\psi_n\rangle|^2$: show that $x=\sum_{n=1}^{\infty} \langle x,\psi_n\rangle \psi_n$
EDIT:
I know I have to show that $\left\|x- \sum_{n=1}^{N} \langle x,\psi_n\rangle \psi_n\rangle\right\|^2 \rightarrow 0 $ as $N \rightarrow \infty $. Then I define $S_n = \sum_{n=1}^{N} \langle x,\psi_n\rangle \psi_n $: so far I have $$ \begin{split} \|x-S_n\|^2 & =\|x\|^2-2\langle x,S_n\rangle+\|S_n\|^2 \\ & = \|x\|^2-2\langle x,S_n\rangle + \Big\| \sum_{n=1}^{N} \langle x,\psi_n\rangle \psi_n \Big\|^2\\ & = \|x\|^2-2\langle x,S_n\rangle + \sum_{n=1}^{N} \|\langle x,\psi_n\rangle \psi_n \|^2\\ & = \|x\|^2-2\langle x,S_n\rangle + \sum_{n=1}^{N} \big(|\langle x,\psi_n\rangle| \cdot \|\psi_n \|\big)^2\\ & = \|x\|^2-2\langle x,S_n\rangle + \sum_{n=1}^{N} \big(|\langle x,\psi_n\rangle|\big)^2 \end{split} $$
For the real case we get $ \langle x, S_n\rangle =\sum\limits_{n=1}^{M}\langle x, \psi_n \rangle =\langle x, \psi_n\rangle=\sum\limits_{n=1}^{M}\langle x, \psi_n \rangle ^{2}$ and you can easily finish the proof from here.
For the complex case, you have made mistake in the computation of $\|x-S_n\|^{2}$. The argument is similar to real case, but you have to remember that inner product is conjugate linear in the second argument.