Given a periodic $f$, we know we can approximate it in $L^2$ with trigonometric polynomials. In particular, $S_n$ converges to $f$ in $L^2$. Now, $\langle \cdot, c \rangle$ is a linear operator, and by the Cauchy-Schwarz inequality, it’s bounded, hence it’s continuous. Thus, by $S_n \to f$, we have $\langle S_n, f \rangle \to \langle f, f \rangle$.
Suppose $H$ is a Hilbert space with basis $\{e_n\}_{n\in \Bbb N}$. Note that
$$\langle f-\langle f, e_k \rangle e_k, \langle f, e_k \rangle e_k \rangle = |\langle f, e_k \rangle |^2 – |\langle f, e_k \rangle|^2 = 0.$$
Therefore $\langle f – S_n, S_n \rangle = 0$. But $\langle f – S_n, S_n \rangle = \langle f, S_n \rangle – \langle S_n, S_n \rangle \to \langle f, f \rangle – \sum_{k\in \Bbb Z} |a_k|^2$. Therefore $\sum_{k \in \Bbb Z} |a_k|^2 = \| f \|^2$.
Alternately, orthogonality implies by the Pythagorean Theorem that $\| f \|^2 = \| f- S_n\|^2 + \|S_n\|^2$. Taking the limit in $n$ gets us the same result.