Let $H$ be a Hilbert space and $\{e_j: j \in I\}$ be a orthonormal basis i.e. $(e_i|e_j) = 0$ whenever $i \neq j$, $||e_j|| = 1$, and the closure of $\text{span}\{e_j:j \in I\}$ is equal to $H$. I know that $\mathbb{F}e_j$ is a closed subspace implying that it's also a Hilbert space. In analysis now, he mentions that $$\sum \mathbb{F}e_j = H$$ where $\sum$ is the direct sum i.e. 0 for all but finitely many elements. I am confused why this is the case.
I tried saying that $x \in H$ i.e. there exists a sequence $\{x_n\}$ in $\text{span}\{e_j:j \in I\}$ that converges to $x$. However, I am having trouble seeing that $x \in \sum \mathbb{F}e_j$.
This statement is false. As has already been commented we can consider the sequence $x_n=\sum_{k=1}^n\frac{1}{k}e_k$. Since the sequence is cauchy it has a limit $x$ and it is easy to see that $<x,e_j>\neq 0$ for infinitely many indices and $x$ can thus not be written as a finite linear combination of $e_j$. It is also easy to see that this subspace of $H$ is in fact equal to $\text{span}\{e_j\}$. The closure of this space is as you know $H$.
What is true however is that every $x$ can be written as a countably infinite linear combination of $e_j$ (even if the basis is uncountable). Since we know that the closure of the span is all of $H$ we can choose a sequence $x_k$ in the span that converges to $x$. Each $x_k$ is a finite linear combination of $e_j$ and so $<x_k,e_j>=0$ for all but finitely many $j$. This implies that $<x,e_j>=\lim_k<x_k,e_j>=0$ for all but countably many $j$. One can now show that $\sum <x,e_j>e_j$ converges to $x$.