Every complete orthonormal set in a Hilbert space $H$ is an orthonormal basis, if and only if $H$ is finite dimensional. Show that any orthonormal set in a Hilbert space $H$ is linearly independent, and use this to show that $H$ is finite dimensional if and only if every complete orthonormal set is an orthonormal basis.
Attempt: 
Trying to show that every orthonormal set is linearly independent is easy if the set is countable, but if it weren't, I'm uncomfortable about considering uncountable sums ($\sum_{\alpha \in I} c_{\alpha}e_{\alpha} = 0$). Anyway around this?
Once I show linear independence, how do prove the claim using it?
 A: As mentioned in the comments, your use of the term "orthonormal basis" is unconventional : Usually a complete orthonormal set is an orthonormal basis. However, if the space is infinite dimensional, then such a set cannot be a Hamel basis :


*

*Any orthonormal set is linearly independent: You don't need uncountable sums here! If $A$ is an orthonormal set and $\{e_1,e_2,\ldots, e_n\} \subset A$ and $\{\alpha_1,\alpha_2,\ldots, \alpha_n\}$ scalars, then
$$
\sum_{i=1}^n \alpha_i e_i = 0 \Rightarrow \alpha_j = \langle \sum_{i=1}^n \alpha_i e_i, e_j\rangle = 0 \quad\forall 1\leq j\leq n
$$
Hence, $A$ is linearly independent.

*If $H$ is a finite dimensional Hilbert space, then every complete orthonormal set has $\dim(H)$ elements. Since they are linearly independent by (1), they must form a Hamel basis.

*If $H$ is infinite dimensional and $A$ is an orthonormal basis, then $A$ must be infinite (as before). So choose a countable subset $\{e_n : n\in \mathbb{N}\} \subset A$. Then the series
$$
\sum_{n=1}^{\infty} \left\| \frac{e_n}{n^2} \right\|
$$
converges. Since $H$ is complete, there is a vector $x\in H$ such that
$$
x = \sum_{n=1}^{\infty} \frac{e_n}{n^2}
$$
Check that $x$ cannot be expressed as a finite linear combination of elements of $A$. Hence, $A$ cannot be a Hamel basis.
