# Dense subset in orthogonal direct sum of Hilbert spaces

I'm reading a proof about orthonormal basis in Hilbert spaces and got confused about a small part.

Let $$\mathcal{H}$$ be a Hilbert space, $$\{ x_i : i \in I \}$$ an orthonormal basis, $$I$$ is an uncountable set.

Denote $$\mathcal{H_i} := \text{span }x_i$$ and let $$\oplus_{i \in I}\mathcal{H_i}$$ be their orthogonal direct sum, i.e

$$x \in \oplus_{i \in I}\mathcal{H_i} \implies x = (\lambda_i x_i)_{i \in I}$$

and $$\{ \|\lambda_ix_i\|^2 : i \in I \}$$ is summable in $$\mathbb{R}$$.

Let $$F$$ denote the linear subspace of all $$(\lambda_ix_i)_{i \in I}$$ with $$\lambda_i \neq 0$$ for at most finite number of indices $$i \in I$$. Then $$F$$ is by definition dense in $$\oplus_{i \in I} \mathcal{H}_i$$

I couldn't convince myself of that. My reasoning is, since $$\{\|\lambda_ix_i\|^2\}_{i \in I}$$ is summable and the orthogonal direct sum of $$1$$-dimensional hilbert spaces is a hilbert space, then $$\{\lambda_ix_i\}_{i \in I}$$ is summable in $$\oplus_{i \in I}\mathcal{H_i}$$. Then $$\lambda_ix_i \neq 0$$ for at most countable number of indices. So somehow, the elements where $$\lambda_ix_i \neq 0$$ for at most finite number of indices can approximate all those elements? How exactly?

Using definition of summability, we can say that $$\big\{\sum_{i \in J} \lambda_ix_i\big\}$$ where $$J \subset I$$ finite, approximate our actual sum $$\sum_{i \in I} \lambda_ix_i$$ arbitrarily well, but I can't see clearly how this implies that $$F$$ is dense.

The partial sums of the series are in $$F$$ and they converge to $$x$$. So every $$x \in \oplus_{i \in I} \mathcal H_i$$ is in the closure of $$F$$.