Showing that the direct sum of pairwise orthogonal and closed vector subspaces is a closed subspace of a Hilbert space $H$ Let $(v_n)_{n \in J}$ be a collection of orthonormal vectors of a Hilbert space $H$, where $J$ is any non-empty index set. I have already shown that the span of each individual vectors is a complete vector subspace and hence closed. I would like to conclude that then the set $S = \bigoplus_{n \in J}\mathrm{span}(v_n)$ is also a closed vector subspace of the ambient Hilber space $H$, but since the number of basis elements is not finite I am not sure how I can achieve this.
For the record, I am interested in showing that the span of the elements $v_n$ is a closed subspace, but as I wasn't quite sure how to work with the potentially ever changing linear combinations (as we might have uncountably many basis vectors), I opted to have a bottom-up approach by 1.) the said direct sum is a closed vector subspace and 2.) showing that the span of all $v_n$s is equal to the said direct sum.
 A: In fact, $S$ is never a closed subspace when $J$ is infinite.

WLOG, suppose $\Bbb N\subseteq J$.
Let $\alpha_n=\sum_{i=1}^n\frac{v_i}i$ for $n\in\Bbb N$. For $m\gt n\ge1$,
$$||\alpha_m-\alpha_n||=||\sum_{i={n+1}}^m\frac{v_n}i||=\left(\sum_{i={n+1}}^m\frac1{i^2}\right)^\frac12<(\frac1{n})^\frac12,$$
where
$$\sum_{i={n+1}}^m\frac1{i^2}<\sum_{i={n+1}}^m\frac1{(i-1)i}=\sum_{i={n+1}}^m(\frac1{i-1}-\frac1{i})=\frac1n-\frac1m<\frac1n.$$
Hence $\{\alpha_n\}_n$ is a cauchy sequence, which converges to an element in $H$. Let it be $\alpha$.

For the sake of contradiction, suppose $\alpha\in S$, i.e., $\alpha=\sum_{i\in J_\alpha}\beta_i$, where $J_\alpha$ is some finite index set, $\beta_i\in \mathrm{span}(v_i)$. Since $\Bbb N$ is infinite, there exists $n_0\in\Bbb N$ such that $n_0\not \in J_\alpha$.
$$\langle \alpha, v_{n_0}\rangle = \langle\lim_{n\to\infty}\alpha_n, v_{n_0}\rangle=\lim_{n\to\infty}\langle\alpha_n, v_{n_0}\rangle=\lim_{n\to\infty}\frac1{n_0}=\frac1{n_0}.$$
On the other hand,
$$\langle \alpha, v_{n_0}\rangle = \langle\sum_{i\in J_\alpha}\beta_i, v_{n_0}\rangle=\sum_{i\in J_\alpha}\langle\beta_i, v_{n_0}\rangle=\sum_{i\in J_\alpha}0=0.$$
This contradiction means $\alpha\not\in S$.
