Variant of Parseval equality in Hilbert space

In fact, you have nearly worked out it. If we choose $q=0$ and $m=q+q'=q'$ in your step, We will have $$\Vert S_{m}\Vert^2 \leqslant \sum_{n=1}^{+\infty}\Vert x_n\Vert^2+2\sum_{n=1}^{m}\Vert x_n\Vert\sum_{j\in I_n}\Vert x_j\Vert$$ And we can estimate the second term: $$\begin{align} 2\sum_{n=1}^{m}\Vert x_n\Vert\sum_{j\in I_n}\Vert x_j\Vert&=\sum_{n=1}^{m}\sum_{j\in I_n}2\Vert x_j\Vert\Vert x_n\Vert\\&\le \sum_{n=1}^{m}\sum_{j\in I_n}(\Vert x_j\Vert^2+\Vert x_n\Vert^2)\\&=\sum_{n=1}^{m}\sum_{n-N_0\le j\le n+N_0}(\Vert x_j\Vert^2+\Vert x_n\Vert^2), \end{align}$$ here $x_j=0$ if $j\le 0$. and we have $$\begin{align} &\sum_{n=1}^{m}\sum_{n-N_0\le j\le n+N_0}(\Vert x_j\Vert^2+\Vert x_n\Vert^2)\\ &\le \sum_{n=1}^{m}(2N_0 \Vert x_n\Vert^2+\sum_{n-N_0\le j\le n+N_0}\Vert x_j\Vert^2)\\ &=2N_0\sum_{n=1}^{m}\Vert x_n\Vert^2+\sum_{n=1}^{m}\sum_{n-N_0\le j\le n+N_0}\Vert x_j\Vert^2\\ &=2N_0\sum_{n=1}^{m}\Vert x_n\Vert^2+\sum_{n=1}^{m}\sum_{|n-j|\le N_0}\Vert x_j\Vert^2\\ &\le 2N_0\sum_{n=1}^{\infty}\Vert x_n\Vert^2+\sum_{n=1}^{\infty}\sum_{j=1}^{\infty}\Vert x_j\Vert^2 I_{\{|n-j|\le N_0\}}\\ &=2N_0\sum_{n=1}^{\infty}\Vert x_n\Vert^2+\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}\Vert x_j\Vert^2 I_{\{|n-j|\le N_0\}} \\ &\le 2N_0\sum_{n=1}^{\infty}\Vert x_n\Vert^2+\sum_{j=1}^{\infty}2N_0\Vert x_j\Vert^2 \\&\le 4N_0\sum_{n=1}^{\infty}||x_n||^2 \end{align} $$ Where $I_{\{|n-j|\le N_0\}}=1$ if $|n-j|\le N_0$,and $0$ otherwise.So we have $$||S_m||^2\le (4N_0+1)\sum_{n=1}^{\infty}||x_n||^2$$ for all $m>0$. Hence $C=4N_0+1$ is the desired constant.

$$||\sum_{i=0}^\infty x_i||^2 = ||\sum_{r = 0}^{N_0 - 1} \sum_{k = 0}^\infty x_{N_0k+r}||^2 \leq \left(\sum_{r=0}^{N_0-1}||\sum_{k=0}^\infty x_{N_0k+r}||\right)^2 \leq N_0\sum_{r=0}^{N_0-1}||\sum_{k=0}^\infty x_{N_0k+r}||^2 $$ $$= N_0 \sum_{i =0}^\infty ||x_i||^2 $$