Define the Fourier coefficients of a function to be
$\hspace{2cm}\mathscr{(F}f)(n)=\frac{1}{2\pi}\int_{0}^{2\pi}f(x)e^{-inx}dx$.
(This is $c_{n}$ in the book, but we indicate explicit dependence on the function.) In this definition the Fourier transform, Parseval's identity reads:
$\hspace{2cm}\int_{0}^{2\pi}\mid f(x) \mid ^2dx=C\sum_{n\in\mathbb{Z}}\mid (\mathscr{F}f(n)\mid ^2$
where C is some absolute positive constant (i.e. independent of $f$).
A $2\pi$-periodic function, $f:\mathbb{T}\rightarrow \mathbb{R}$ is square summable if
$\hspace{2cm}\int_{0}^{2\pi}\mid f(x)\mid^2dx<\infty$
Use Pareval's identity to show that for some (possibly different) absolute constant C, and real-valued, $2\pi$ -periodic, square summable functions $f$ and $g$,
$\hspace{2cm}\sum_{n}\Re((\mathscr{F}f)(n)(\mathscr{F}g)(-n))=C\int_{0}^{2\pi}f(x)g(x)dx$
By the identical polarization in the norm identity $$ \frac1{2\pi}\int_0^{2\pi}|f(x)|^2\,dx=\|f\|_{[0,2\pi]}=\|\F f\|_{\ell^2(\Bbb Z)}=\sum_{n\in\Bbb Z}|(\F f)(n)|^2 $$ to get the Euclidean scalar products (interpreting the complex Hilbert spaces as real vector spaces) $$ ⟨a,b⟩=\frac14\Bigl(\|a+b\|^2-\|a-b\|^2\Bigr) $$ you get $\newcommand{\F}{\mathcal{F}}$ $$ \Re\left(\frac1{2\pi}\int_0^{2\pi}f(x)\overline{g(x)}\,dx\right) =\langle f,g\rangle_{[0,2\pi]} =⟨\F f,\F g⟩_{\ell^2(\Bbb Z)} =\Re\left(\sum_{n\in\Bbb Z}(\F f)(n)\overline{(\F g)(n)}\right). $$ Now use that directly from the integral definition of the Fourier coefficients $$ (\F g)(-n)=\overline{(\F\bar g)(n)} $$ to get to the conclusion.